When you need to construct a circle, the major factor to consider is the radius.
The radius is the same distance from any point around the circumference of the circle to the centre. Since the radius is not given, you however need to look for clues.
You start by joining the points to arrive at two lines, for example, join points X and Y and then join points Y and Z.
Next you bisect each of the two lines one after the other (bisect along the perpendicular)
You will observe that both perpendicular bisectors would touch at a point. That point where they touch or "cross each other" is the center of your circle.
Next you place the sharp tip of your compass on the center of your circle, adjust its distance to the pencil end (that is your radius) and as soon as it touches one of the three points, you draw your circle.
8. MOVIE TICKETS Tickets to a movie cost $25 for adults and 5.50 formodents A group of friends purchased 8 tickets for $52.75 a Write a system of equations to represent the station
Tickets for adults --> $25
Tickets for formodents --> $5.50
The equations that would represent the number of adults and formodents in th group of friends:
Let x be adults
Let y be formodents
$25x+$5.50y=$52.72
x+y=8
If the sum of a number and nine is tripled , the result is two less than twice the number. Find the number.
By solving a simple linear equation we will see that the number is -29.
How to find the number?Let's define x as the number, then the sentence:
"If the sum of a number and nine is tripled , the result is two less than twice the number"
Can be written as the equation:
3*(x + 9) = 2x - 2
This is a linear equation that we can solve for x:
3*(x + 9) = 2x - 2
3x + 27 = 2x - 2
3x - 2x = -2 - 27
x = -29
The number is -29.
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7. A large cooler contains the following drinks: 5 lemonades, 9 Sprites, 7 Cokes, and 10 root beers. You randomly pick two cans, one at a time (without replacement). Compute the following probabilities.(a) What is the probability that you get two cans of Sprite? (b) What is the probability that you do not get two cans of Coke? (c) What is the probability that you get either two root beers or two lemonades? (d) What is the probability that you get one can of Coke and one can of Sprite? (e) What is the probability that you get two drinks of the same type?
A large cooler contains the following drinks,
5 Lemonades, let L reprersent Lemonade
9 Sprites, let S reprersent Sprite
7 Cokes, let C represent Coke and
10 Root beers, let R represent Root beer
Total drinks in the cooler is
[tex]=5+9+7+10=31[/tex]Total outcome = 31 drinks
The formula of probability is
[tex]\text{Probability}=\frac{required\text{ outcome}}{total\text{ outcome}}[/tex]a) The probability that you get two cans of Sprite is
[tex]\begin{gathered} Prob\text{ of picking the first Sprite without replacement is} \\ P(S_1)=\frac{9}{31} \\ Prob\text{ of picking the second Sprite is} \\ P(S_2)=\frac{8}{30} \\ \text{Probability of getting two cans of Sprite}=(PS_1S_2)=\frac{9}{31}\times\frac{8}{30}=\frac{12}{155} \\ (PS_1S_2)=\frac{12}{155} \end{gathered}[/tex]Hence, the the probability that you get two cans of Sprite is 12/155
b)
The probability that you get two cans of Coke
[tex]\begin{gathered} Prob\text{ of picking the first Coke without replacement is} \\ P(C_1)=\frac{7}{31} \\ Prob\text{ of picking the second Coke is} \\ P(C_2)=\frac{6}{30} \\ \text{Probability of getting two cans of Coke is} \\ P(C_1C_2)=\frac{7}{31}\times\frac{6}{30}=\frac{7}{155} \\ P(C_1C_2)=\frac{7}{155} \end{gathered}[/tex]The probability that you do not get two cans of Coke will be
[tex]\begin{gathered} \text{Prob that you do not get two cans of Coke is} \\ 1-P(C_1C_2)=1-\frac{7}{155}=\frac{155-7}{155}=\frac{148}{155} \\ \text{Prob that you do not get two cans of Coke }=\frac{148}{155} \end{gathered}[/tex]Hence, the probability that you do not get two cans of Coke is 148/155
c)
The probability that you get two cans of Root beers is
[tex]\begin{gathered} Prob\text{ of picking the first Root b}eer\text{ without replacement is} \\ P(R_1)=\frac{10}{31} \\ Prob\text{ of picking the second Root b}eer\text{ is} \\ P(R_2)=\frac{9}{30} \\ \text{Probability of getting two cans of Root b}eer\text{ is} \\ P(R_1R_2)=\frac{10}{31}\times\frac{9}{30}=\frac{3}{31} \\ P(R_1R_2)=\frac{3}{31} \end{gathered}[/tex]The probability that you get two cans Lemonades is
[tex]\begin{gathered} Prob\text{ of picking the first Lemonade without replacement is} \\ P(L_1)=\frac{5}{31} \\ Prob\text{ of picking the second Root b}eer\text{ is} \\ P(L_2)=\frac{4}{30} \\ \text{Probability of getting two cans of Lemonade is} \\ P(L_1L_2)=\frac{5}{31}\times\frac{4}{30}=\frac{2}{93} \end{gathered}[/tex]The probability that you get either two root beers or two lemonades is
[tex]P(R_1R_2)+P(L_1L_2)=\frac{3}{31}+\frac{2}{93}=\frac{11}{93}[/tex]Hence, the probability that you get either two root beers or two lemonades is 11/93
d)
[tex]\begin{gathered} Prob\text{ of picking the first Coke without replacement is} \\ P(C)=\frac{7}{31} \\ \text{Prob of picking a can of Sprite is} \\ P(S)=\frac{9}{30} \end{gathered}[/tex]After getting both Sprite and Coke you will multiply the probabilities and then multiply them with 2 because you may choose Coke in first try and Sprite in second or the other way around
The probability that you get one can of Coke and one can of Sprite is
[tex]P(CandS)=2\times(\frac{7}{31}\times\frac{9}{30})=2(\frac{21}{310})=\frac{21}{155}[/tex]Hence, the probability that you get one can of Coke and one can of Sprite is 21/155
e)
Prob of two of each of the cans of drinks (without replacement) are as follow
[tex]\begin{gathered} P(L_1L_2)=\frac{2}{93} \\ (PS_1S_2)=\frac{12}{155} \\ P(C_1C_2)=\frac{7}{155} \\ P(R_1R_2)=\frac{3}{31} \end{gathered}[/tex]The probability that you get two drinks of the same type is
[tex]\begin{gathered} \text{Prob of two drinks of the same type is} \\ =P(L_1L_2)+(PS_1S_2)_{}+P(C_1C_2)+P(R_1R_2) \\ =\frac{2}{93}+\frac{12}{155}+\frac{7}{155}+\frac{3}{31}=\frac{112}{465} \\ \text{Prob of two drinks of the same type}=\frac{112}{465} \end{gathered}[/tex]Hence, the probability that you get two drinks of the same type is 112/465
please explain What is the simplified form of the expression?
2x 2 + 4y + 3x 2 – 2y + 3y
The simplified form of the expression is found to be 5(x² + y) by adding or subtracting all the similar terms
What is the difference between a mathematical expression and an equation?A number, a variable, or a mix of numbers, variables, and operation symbols make up an expression. Two expressions joined by an equal sign form an equation.
What does "simplification of an algebraic expression" mean?The technique of expressing an algebraic expression in the most effective and compact form without altering the original expression's value is known as simplification. The procedure involves gathering related terms, which calls for adding or removing terms from an expression.
The given expression is 2x² + 4y + 3x² -2y +3y
We need to simplify this expression.
2x² + 4y + 3x² -2y +3y
= 2x² + 3x² + 4y - 2y + 3y
= 5x² + 5y
= 5(x² + y)
Therefore, the simplified form of the expression is found to be 5(x² + y) by adding or subtracting all the similar terms.
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the parent function name for y=|x|
This function is an absolute type of function
7(-a-3)=3(2a-6) I have the answer but I need help checking it.
SOLUTION:
Step 1:
In this question, we are meant to solve the following:
[tex]7\text{ ( - a - 3 ) = 3 ( 2a - 6 )}[/tex]Step 2:
Simplifying, we have that:
[tex]\begin{gathered} -7a\text{ - 21 = 6a - 18} \\ \end{gathered}[/tex]collecting like terms, we have that:
[tex]\begin{gathered} -21\text{ + 18 = 6 a + 7a} \\ 13\text{ a = -3} \\ \text{Divide both sides by 13, we have that:} \\ a\text{ = }\frac{-3}{13} \end{gathered}[/tex]Step 3:
To verify that:
[tex]a\text{ =}\frac{-3}{13}[/tex]is a solution, we have that:
[tex]7\text{ ( - a - 3 ) = 7 \lbrack -(}\frac{-3}{13}\text{ ) - 3 \rbrack}[/tex][tex]7\lbrack\text{ }\frac{3}{13}\text{ - 3\rbrack = 7 \lbrack }\frac{3}{13}\text{ - }\frac{39}{13}\text{ \rbrack = 7 x }\frac{-36}{13}\text{ = }\frac{-252}{13}\text{ ( Left Hand Side)}[/tex]Next,
[tex]3\text{ ( 2 a - 6 ) = 3 \lbrack{}2(}\frac{-3}{13})\text{ - 6 }\rbrack\text{ = 3 \lbrack}\frac{-6}{13}\text{ - 6\rbrack= 3\lbrack}\frac{-6}{13}\text{ - }\frac{78}{13}\rbrack[/tex][tex]=\text{ 3 \lbrack }\frac{-84}{13}\text{ \rbrack = }\frac{-252}{13}\text{ ( Right Hand Side)}[/tex]CONCLUSION:
From the solution and from the verification of the answers, we can see that the correct answer is:
[tex]a\text{ = }\frac{-\text{ 3}}{13}[/tex]Simplify f(x) = 2x^5 for x = 0, 1, 3, 5
f(0) = 0, f(1) = 2, f(3) = 486, f(5) = 6250
Explanations:The given function is:
[tex]f(x)=2x^5[/tex]To get the value of f(x) for x = 0, 1, 3, and 5, it means we are going to find f(0), f(1), f(3), and f(5).
[tex]\begin{gathered} f(0)=2(0)^5 \\ f(0)\text{ = 2(0)} \\ f(0)\text{ = 0} \end{gathered}[/tex][tex]\begin{gathered} f(1)=2(1)^5 \\ f(1)\text{ = 2(1)} \\ f(1)\text{ = 2} \end{gathered}[/tex][tex]\begin{gathered} f(3)=2(3)^5 \\ f(3)\text{ = 2 (}243) \\ f(3)\text{ = 486} \end{gathered}[/tex][tex]\begin{gathered} f(5)=2(5)^5 \\ f(5)\text{ = 2(}3125) \\ f(5)\text{ = }6250 \end{gathered}[/tex]Find the indicated probability. Round your answer to 6 decimal places when necessary.Find the probability of tossing 1 tails or 1 head on the first 8 tosses of a "fair" coin.
First, find the probability of getting 1 head in 8 tosses
P(1 head) = tosses with exactly 1 head/ total number of possible outcomes
number of outcomes = 2^8 = 256
number of outcomes with 1 head = 8 ( we could get 1 head on the first toss or on the second or on the third......)
P (1 head) = 8/256 = 1/32
The same would be true for tails
P(1 tail) = tosses with exactly 1 tail/ total number of possible outcomes
= 8/256 = 1/32
The formula to calculate the “or” probability of two events A and B is this: P(A OR B) = P(A) + P(B) – P(A AND B).
Since we cannot get P(1 head and 1 tail) since we toss 8 times
P (1head or 1 tail) = 1/32 + 1/32 = 2/32 = 1/16 =.0625
Find the x-intercept and the y-intercept without graphing. Write the coordinates of each intercept. When typing the point (x,y) be sure to include parentheses and a comma between your x and y components. Do not put any spaces between your characters. If a value is not an integer type your answer rounded to the nearest hundredth.3x+8y=24the x-intercept is Answerthe y-intercept is Answer
We want to find the x and y-intercepts of
[tex]3x+8y=24[/tex]The x-intercept is where the graph cuts the x-axis, when y = 0. To find this in our equation, we just need to evaluate it at y = 0.
[tex]\begin{gathered} 3x+8\times0=24 \\ 3x=24 \\ x=\frac{24}{3}=8 \end{gathered}[/tex]Then, the x-intercept is (8, 0).
The y-intercept is where the graph cuts the y-axis, when x = 0. To find this in our equation, we just need to evaluate it at x = 0.
[tex]\begin{gathered} 3\times0+8y=24 \\ 8y=24 \\ y=\frac{24}{8}=3 \end{gathered}[/tex]The y-intercept is (0, 3).
what is the range of the function graphed below?[tex]1 \leqslant y \ \textless \ 4 \\ - 3 \ \textless \ y \leqslant 3 \\ - 2 \leqslant y \leqslant 3 \\ - 3 \leqslant y \ \textless \ 4[/tex]
The range of the function is (-3, 3]
The range of a function is composed by all the values that y reaches in the function. Here we can see that the functions goes from 3 to -3. Then the range set is (-3, 3]. It has a parentheses in -3 because the function doesn't reach -3
1f(x) =X-24g(x)ХFind: (fog)(x) =
We have the functions:
[tex]undefined[/tex]s
Write the first five terms of each sequence a(1) = 7, a(n) = a(n - 1) - 3 for n = 2.
Answer:
7 , 4, 1, -2 and -5
Explanation:
Given a sequence such that:
[tex]\begin{gathered} a(1)=7 \\ a(n)=a(n-1)-3,n\geqslant2 \end{gathered}[/tex][tex]\begin{gathered} a\left(2\right)=a\left(2-1\right)-3=a(1)-3=7-3=4\implies a(2)=4 \\ a\left(3\right)=a\left(3-1\right)-3=a(2)-3=4-3=1\implies a(3)=1 \\ a\left(4\right)=a\left(4-1\right)-3=a(3)-3=1-3=-2\implies a(4)=-2 \\ a\left(5\right)=a\left(5-1\right)-3=a(4)-3=-2-3=-5\implies a(5)=-5 \end{gathered}[/tex]Therefore, the first five terms of the sequence are:
7 , 4, 1, -2 and -5
what is the equation of a line that passes through point (-1,5) and has the slope of m=4
The general equation of a line is given as;
[tex]y=mx+b[/tex]In this question, the slope (which is m) is given as 4. Also we have the points x and y, given as (-1, 5). That is;
[tex]x=-1,y=5[/tex]Therefore the next step is to find the y-intercept (that is b in the equation).
We substitute for the known values as follows;
[tex]\begin{gathered} y=mx+b \\ 5=4(-1)+b \\ 5=-4+b \\ \text{Add 4 to both sides} \\ 5+4=-4+4+b \\ 9=b \end{gathered}[/tex]Now we know the value of b and m, we can substitute them as follows;
[tex]\begin{gathered} y=mx+b \\ m=4,b=9 \\ y=4x+9 \end{gathered}[/tex]A state sales tax of 6% and a local sales tax of 1% are levied in Tampa, Florida. Suppose the price of a particular car in Tampa is $15,000, and an oil change at a certain auto center is $29.Which statement is true another total cost of the car and the oil change after sales tax has been calculated?Select the correct answer
We have the following:
What we must do is calculate the total cost of the car by adding its original value plus the cost of taxes, 6% and 1%
We know that the initial value is $15000, if to that we add 6% of those $15000 and equal 1%, we have
[tex]15000+15000\cdot0.06+15000\cdot0.1=15000+900+150=16050[/tex]We do the same procedure for the oil change
[tex]29+29\cdot0.06+29\cdot0.01=29+1.74+0.29=31.03[/tex]Therefore the correct statement is the last
What is the 15th term in the sequence using the given formula?
Solution:
The formula is given below as
[tex]c_n=3n-1[/tex]Concept:
To figure out the 15th term, we will substitute n=15
By substituting values, we will have
[tex]\begin{gathered} c_{n}=3n-1 \\ c_{15}=3(15)-1 \\ c_{15}=45-1 \\ c_{15}=44 \end{gathered}[/tex]Hence,
The final answer is
[tex]\Rightarrow44[/tex]The THIRD OPTION is the right answer
Given the kite ABCD, which statement is false?Just the answer.
Explanation
let's check every option
then
A)
[tex]\angle ADC\text{ is congruente to }\angle ABC[/tex]we can see those angles (in black), aand as the lengths of the sides are similar this angles are congruente,so
[tex]\begin{gathered} \angle ADC\text{ is congruent to }\angle ABC \\ \text{true} \end{gathered}[/tex]B)
[tex]undefined[/tex]help with a ab math question
It seems to be a technical issue with the tool
I can't open the image
Amanda likes to launch model rockets. For one of Amanda's rockets, the function S(t)= −16t^2+41t+112 gives the height of the rocket above the ground in feet, in terms of the number of seconds t since the rocket's engine stops firing.Please use 4 or more decimals.How far above the ground is the rocket when it stops firing?After how many seconds does the rocket reach its maximum height?What is the maximum height reached by the rocket?After how many seconds will the rocket hit the ground?
Answer:
• (a)112 feet
,• (b)1.28125 seconds.
,• (c)138.265625 feet.
,• (d)4.22091 seconds
Explanation:
The height of the rocket in terms of the number of seconds t since the rocket's engine stops firing is given below.
[tex]S\mleft(t\mright)=-16t^2+41t+112[/tex]Part A
At the time the rocket stopped firing, t=0.
[tex]S(0)=-16(0)^2+41(0)+112=112[/tex]The rocket was 112 feet above the ground when it stopped firing.
Part B
The value of t at which the rocket reaches its maximum height is the equation of the line of symmetry.
To find this equation, we use the formula below.
[tex]t=-\frac{b}{2a}=-\frac{41}{-2\times16}=1.28125\text{ seconds}[/tex]The rocket reaches its maximum height after 1.28125 seconds.
Part C
To find the maximum height, substitute t=1.28125 into S(t).
[tex]\begin{gathered} S\mleft(t\mright)=-16t^2+41t+112 \\ \implies S(1.28125)=-16(1.28125)^2+41(1.28125)+112 \\ =138.265625\text{ ft} \end{gathered}[/tex]The maximum height of the rocket is 138.265625 feet.
Part D
When the rocket hits the ground, the height is 0.
Set S(t)=0 and solve for t as follows.
[tex]S(t)=-16t^2+41t+112=0[/tex]Using the quadratic formula:
[tex]\begin{gathered} t=\dfrac{-41\pm\sqrt[]{41^2-4(-16)(112)}}{2\times-16}=\dfrac{-41\pm\sqrt[]{1681-(-7168)}}{-32} \\ =\dfrac{-41\pm\sqrt[]{1681+7168}}{-32} \\ =\dfrac{-41\pm\sqrt[]{8849}}{-32} \\ t=\dfrac{-41+\sqrt[]{8849}}{-32}\text{ or }t=\dfrac{-41-\sqrt[]{8849}}{-32} \\ t=-1.658\; \text{or }t=4.22091 \end{gathered}[/tex]Since t cannot be negative, the rocket will hit the ground after 4.22091 seconds.
I need help on this equation. It’s algebra. SAT PREP.
Answer:
r = 1.14
Explanation:
The value of a product (A) over time with an increasing rate "i" can be calculated as follows:
A = C(1+i)^t
where:
C is the value of the product at time 0;
A is the value of the product at time t;
i is the increasing rate.
If we compare the expression V=300r^t with A = C(1+i)^t. We can observe that:
r = 1+i
r = 1+0.14
r = 1.14
Really need help solving this practice from my ACT prep guide It’s a trig practice
Given:
- The amplitude of the Sine Function:
[tex]A=10[/tex]- The midline:
[tex]y=4[/tex]- And the period:
[tex]Period=2[/tex]- You know that the function does not have a Phase shift.
• You need to remember that, by definition, the General Equation for a Sine Function has this form:
[tex]y=Asin\mleft(B\mleft(x+C\mright)\mright)+D[/tex]Where "A" is the amplitude, "C" is the phase shift, "D" is the vertical shift and this is the period:
[tex]Period=\frac{2\pi}{B}[/tex]Since the midline is given by the vertical shift, you can identify that, in this case:
[tex]D=4[/tex]And, knowing the period, you can set up that:
[tex]2=\frac{2\pi}{B}[/tex]Solving for "B", you get:
[tex]\begin{gathered} 2B=2\pi \\ \\ B=\frac{2\pi}{2} \\ \\ B=\pi \end{gathered}[/tex]• It is important to remember the following Transformation Rule for Functions:
When:
[tex]-f(x)[/tex]The function is reflected over the x-axis.
Therefore, knowing all the data, you can set up this equation:
[tex]f(x)=-10\sin (\pi x)+4[/tex]Hence, the answer is: First option.
could someone help me find the measures of this Rhombus? im very confused right now and need an explanation on thisThe measures you need to find:NK=NL=ML=JM=M
We shall take a quick reminder of the properties of a rhombus.
All sides are equal in measure
The opposite sides are parallel
The diagonals bisect each other at right angles
Opposite angles are equal in measure
Therefore, we can deduce the following from the given rhombus;
If JL bisects MK, then
[tex]\begin{gathered} MN=NK=\frac{MK}{2} \\ MN=NK=\frac{24}{2} \\ MN=NK=12 \end{gathered}[/tex]If MK bisects JL, then line
[tex]\begin{gathered} JN=NL=\frac{JL}{2} \\ JN=NL=\frac{20}{2} \\ JN=NL=10 \end{gathered}[/tex]Also, in triangle MJN,
MN = 12,
JN = 10,
Angle J = 50
Angle N = 90
Therefore angle M = 40
(All three angles in a triangle sum up to 180)
Therefore, in right angled triangle MJN, with the right angle at N,
[tex]\begin{gathered} MN^2+JN^2=JM^2 \\ 12^2+10^2=JM^2 \\ 144+100=JM^2 \\ 244=JM^2 \\ \sqrt[]{244}=JM \\ JM=15.6 \end{gathered}[/tex]All sides are equal, therefore,
JM = ML = 15.6
Since line MK has been bisected by line JL, then
[tex]\angle KNL=90[/tex]Also angle MJL equals 50, and line JL bisects angle J, then
[tex]\angle MJL=\angle KJL=50[/tex]If angle MJL and angle KJL both measure 50, then angle MJK equals 100 (50 + 50).
Opposite angles of a rhombus are equal, hence
[tex]\angle MJK=\angle MLK=100[/tex]If KJL = 50, and JNK = 90, then
[tex]\begin{gathered} \angle JKM+\angle KJL+\angle JNK=180\text{ (angles in a triangle sum up to 180)} \\ \angle JKM+50+90=180 \\ \angle JKM=180-50-90 \\ \angle JKM=40 \end{gathered}[/tex]If JKM = 40, then
[tex]\begin{gathered} \angle JKM=\angle LKM=40 \\ \angle JKL=\angle JKM+\angle LKM \\ \angle JKL=80 \\ \angle JKL\text{ and }\angle JML\text{ are opposite angles. Therefore,} \\ \angle JML=80 \end{gathered}[/tex]So the answers are;
NK = 12
NL = 10
ML = 15.6
JM = 15.6
How many different regrestation codes are possible. And also what is the probability that all the first three digits of the code are not even numbers.
a) Consider the 7-digit registration code to be an arrangement of 7 cells to be filled using the given digits.
In the first cell, one can write any of the digits; on the other hand, there are only 6 digits available to fill the second cell (no number can be used more than once). Therefore, there are 5 digits that can be used in the third cell and so on; thus, there is a total of
[tex]7*6*5*4*3*2*1=7!=5040[/tex]5040 different registration codes.b) The 5040 different combinations found above are equally probable.
There are only 3 available even numbers (2, 4, and 6); therefore, we need to find the number of combinations such that none of the first three digits is equal to 2, 4, or, 6.
Thus, using a diagram,
There are 4 possible numbers that one can fit in the first cell (1,5,7, or 9), in the second cell, one can fit 3 numbers (any of the remaining ones from cell 1), and so on.
In the fourth cell (first cell in blue), one can fit any even number plus a remaining odd number from cell 3.
Therefore, the total number of codes such that their first three digits are not even are
[tex]4*3*2*4*3*2*1=576[/tex]Then, the corresponding probability is
[tex]P=\frac{576}{5040}=\frac{4}{35}[/tex]The answer to part b) is 4/35For the problem below, find the reference angle, to the nearest 10th (if necessary), and also the two possible quadrants in which θ could lie.tan(θ)=−3
The two possible quadrants are the second and the fourth
5) Francisco practiced playing his violin for 2 1/3 hours on Sunday. He practiced for 5/6 hour on Monday. How much time did Francisco spend playing his violin?(C)1 hours 3 (A)1 hours (B) hour (D) 3-hours, 10 min
Answer:
D
Francisco spent 3 hours, 10 minutes playing his violin
Explanation:
Given that:
Francisco practised playing his violin for
2 hours on Sunday
5/6 hours on Monday
The total number of time he spends playing his violin is obtained by adding the number of hours he spends each day.
[tex]\begin{gathered} 2\frac{1}{3}+\frac{5}{6} \\ \\ =\frac{7}{3}+\frac{5}{6} \\ \\ =\frac{19}{6} \\ \\ =3\text{ }\frac{1}{6} \end{gathered}[/tex]This is 3 hours, 10 minutes.
If 2 dogs cross over a road and 1 dog disappear in the road how did the other dog made it
Answer:
he was quick???
Step-by-step explanation:
Mark me brainliest!
which of the following could be the combination of values for the students and the minimum numbers of chaperones the museum requires
3 chaperones ---------------------------- 24 students
9 chaperones --------------------------- 72 students
2chaprones 16 students
7.5 chaperones ----------------------- 60 students
5.6 chaperones------------------------ 45 students
5 chaperones ------------------------- 40
The first two options are correct
10x2+4x factor completely
Answer:
2x * ( 5x + 2 )
Step-by-step explanation:
10x^2 = 2x * 5x
4x = 2x * 2
10x^2 + 4x = 2x * ( 5x + 2 )
-Quadratic Equations- Determine the number and the nature of the solutions to (3a + 24)² = -36 and then solve
ANSWER
There are two solutions and they are both complex solutions. The solutions are:
[tex]a=2i-8;a=-2i-8[/tex]EXPLANATION
We want to determine the number and nature of solutions to the equation:
[tex](3a+24)^2=-36[/tex]To do this, solve the equation by first, finding the square root of both sides of the equation:
[tex]\begin{gathered} \sqrt[]{(3a+24)^2}=\pm\sqrt[]{-36}=\pm\sqrt[]{-1\cdot36} \\ \Rightarrow3a+24=\pm\mleft\lbrace\sqrt[]{36}\cdot\sqrt[]{-1}\mright\rbrace \\ 3a+24=\pm6i \end{gathered}[/tex]Now, solve the equation for a:
[tex]\begin{gathered} 3a=\pm6i-24 \\ \Rightarrow a=\pm\frac{6i}{3}-\frac{24}{3} \\ \Rightarrow a=2i-8;a=-2i-8 \end{gathered}[/tex]Hence, there are two solutions and they are complex solutions.
Select all the pairs that represent alternate interior angles.See image for instruction
Alternate means on the opposite side of the transversal, or line n
interior means inside of the parallel lines l and m
The alternate interior angles are 4 and 5
and 3 and 8
Check the boxes for both pairs
6. Cesium-137 has a half-life of 30 years. Suppose a lab stores 30 mg in 1975. How much would be left in 2065? y = a (1 + r) (Fill in answer choices for a, r and t.)
The formula for calculating the amount remaining after a number of half years , n is :
[tex]\begin{gathered} A=\frac{A_{\circ}}{2^n^{}} \\ \text{where A}_{\circ}\text{ =initial }amount \\ n=\frac{t}{t_{\frac{1}{2}}} \end{gathered}[/tex]The lab store mass of Cesium-137 is 30mg in 1975
then the mass of Cesium-137 in 2065,
Time period =2065-1975
time period t=90 years,
substitute the value and solve for A
[tex]\begin{gathered} A=\frac{30}{2^{\frac{90}{45}}} \\ A=\frac{30}{2^2} \\ A=\frac{30}{4} \\ A=7.5\text{ mg} \end{gathered}[/tex]In 2065, the mass of Cesium -137 will be 7.5 mg
Answer : 7.5mg