Solve the following system of equations;
[tex]\begin{gathered} 2y=4x+12---(1) \\ y=2x-6---(2) \\ \text{From equation (2) substitute for y=2x-6 into equation (1) } \\ 2(2x-6)=4x+12 \\ 4x-12=4x+12 \\ \text{Collect all like terms} \\ 4x-4x=12+12 \\ 0=24 \end{gathered}[/tex]The answer is 0 = 24, which is not possible.
Hence, the system of equations has NO SOLUTION
Convert percent 26% of a number is what fraction of that number
Express 26% as a fraction:
26% = 26/100
[tex]\frac{26}{100}=\frac{13}{50}=0.26[/tex]The number a exceeds the number b by 50% by what percent is the number b smaller than the number a
Number a exceeds the number b by 50%:
[tex]\begin{gathered} a=b+0.5b \\ a=1.5b=\frac{3}{2}b \\ b=\frac{2}{3}a \end{gathered}[/tex]That means, 1/3 smaller=33, 1/3% smaller or 33% smaller.
16. The graph shows the relationship between the total cost and the amount of rice purchased 32 20 Total price ($) Amount of rice (lb) Part A: What does the ordered pair (6, 30) represent? Part B: Which point on the graph represents the unit price? Part C: How many pounds would you have to buy for the total cost to be $20? Explain how to find the answer
For the information given in the statement you have:
Part A: According to the graph, the point (6,30) means that the total cost of 6 pounds of rice is $30.
Part B: The unit price per pound of rice can be seen in the graph at point (1,4), that is, the price of 1 pound of rice is $ 4.
Part C: To find out how many pounds you would have to buy for the total cost to be $ 20 you have to find the point whose second coordinate is 20, that is, point (4,20), then you would have to buy 4 pounds of rice for the total cost to be $20.
I don't understand how to do a certain equation i have no clue what its called
Answer:
y =24 -x
Explanation:
Mr Ledger has 24 donuts and if he uses x donuts then the amount he will have left will be
[tex]24-x[/tex]and since we are calling this amount left y, we can say
[tex]y=24-x[/tex]which is our answer!
When Mr Ledger uses 18 donuts, the number left will be
[tex]\begin{gathered} y=24-18, \\ y=6. \end{gathered}[/tex]And when Mr Ledger uses 22 donuts, we will have left
[tex]\begin{gathered} y=24-22, \\ y=2. \end{gathered}[/tex]donuts.
The graph of the two points is given below.
write the number in standard notation:3.92x10^7
We will write it as follows:
[tex]3.92\cdot10^7=39200000[/tex]Given the triangle congruence statement ΔUVW≅ΔABC and the triangles below, mark each of the triangles appropriately for corresponding angles and sides. Then create a list of congruent sides and angles using the tableCorresponding SidesCorresponding Angles
Given:
[tex]\Delta UVW\cong\Delta ABC[/tex]Corresponding sides are:
[tex]\begin{gathered} UV\cong AB \\ VW\cong BC \\ UW\cong AC \end{gathered}[/tex]Corresponding angles are:
[tex]\begin{gathered} \angle U\cong\angle A \\ \angle V=\angle B \\ \angle W=\angle C \end{gathered}[/tex]Hello, I'm finding this a tad bit difficult.A little help please. Question 1a. Calculate the total are including the frame.Question 1b: Calculate the external perimeter of this picture frame.
Part a.
In the first part of the problem we need to calculate the total area including the frame, for this, we use the formula for the area of a parallelogram:
In this case, the values of a, b and c are:
[tex]\begin{gathered} a=42\operatorname{cm} \\ b=65\operatorname{cm} \\ c=0.39m=39\operatorname{cm} \end{gathered}[/tex]Thus, the area is:
[tex]\begin{gathered} A=65\operatorname{cm}\times39\operatorname{cm} \\ A=2,535\operatorname{cm}^2 \end{gathered}[/tex]2,535 centimeters squared.
Part b.
In this part, we are asked to find the external perimeter of the picture frame.
For this, we use the formula to find the perimeter of a parallelogram:
Substituting the values of a and b from part a into the perimeter formula:
[tex]\begin{gathered} P=2(a+b) \\ P=2(42\operatorname{cm}+65\operatorname{cm}) \\ P=2(107\operatorname{cm}) \\ P=214\operatorname{cm} \end{gathered}[/tex]The perimeter of the frame is 214 centimeters.
Answer:
A. Area
[tex]2,535\operatorname{cm}^2[/tex]B. Perimeter
[tex]214\operatorname{cm}[/tex]Find the value of X in the length of MO
We are given that:
N is between M & O
[tex]\begin{gathered} MN=2x+4 \\ MO=6x \\ NO=28 \\ \\ \text{If N is the midpoint of M \& O, we have:} \\ MN=NO \\ 2x+4=28 \\ \text{Subtract ''4'' from both sides, we have:} \\ 2x=28-4 \\ 2x=24 \\ \text{Divivde both sides by ''2'', we have:} \\ x=\frac{24}{2}=12 \\ x=12 \\ \\ \therefore x=12 \end{gathered}[/tex]We will obtain the value of MO by substituting the value of ''x'' into MO. We have:
[tex]\begin{gathered} MO=6x \\ x=12 \\ MO=6(12)=72 \\ MO=72 \\ \\ \therefore MO=72 \end{gathered}[/tex]Complete the square to findthe vertex of this parabola.x² - 2x + y - 4 = 0([?], [ ])
Given:
[tex]x^2-2x+y-4=0[/tex]Let's complete the square to find the vertex of the parabola.
To solve first move all terms not containing y to the right side of the equation:
[tex]y=-x^2+2x+4[/tex]Now, take the vertex form of a parabola:
[tex]y=a(x-h)^2+k[/tex]Apply the standard form of a parabola:
[tex]\begin{gathered} ax^2+bx+c \\ \\ -x^2+2x+4 \end{gathered}[/tex]Thus, we have:
a = -1
b = 2
c = 4
Now, to find the value of h, we have:
[tex]\begin{gathered} h=-\frac{b}{2a} \\ \\ h=-\frac{2}{2(-1)} \\ \\ h=-\frac{2}{-2} \\ \\ h=1 \end{gathered}[/tex]To find the value of k, we have:
[tex]\begin{gathered} k=c-\frac{b^2}{4a} \\ \\ k=4-\frac{2^2}{4(-1)} \\ \\ k=4-\frac{4}{-4} \\ \\ k=4+1 \\ \\ k=5 \end{gathered}[/tex]We have the values:
h = 1
k = 5
The vertex of the parabola is:
(h, k) ==> (1, 5)
ANSWER:
(1, 5)
Solve each equation for the variable. -10 = Зm +5
-10 = Зm +5
Subtract 5 from both sides of the equation:
-10-5 = 3m+5-5
-15 = 3m
Divide both sides by 3
-15/3 = 3m/3
-5= m
m= -5
A fireworks rocket is launched from a hill above a lake. The rocket will fall into the lake after exploding in the air. The rocket’s height above the surface of the lake is given by the function g(x)= -16x2 + 64x + 80 where x is the number of seconds after the rocket is launched. The function can also be written in factored form as g(x) = -16 (x + 1)(x - 5).When does the rocket hit the ground?
We are giving the function as;
[tex]g(x)=-16x^2+64x+80[/tex]If we factorize g(x) we have that
[tex]g(x)=-16(x+1)(x-5)[/tex]T0 find if the object has be launched and find the x value of the object
Therefore,
[tex]\begin{gathered} -16(x+1)(x_{}-5)=0 \\ \Leftrightarrow(x+1)(x_{}-5)=0 \\ \leftrightarrow x=-1\text{ or x=5} \end{gathered}[/tex]it will take the rocket 5 seconds to reach the ground.
Sarah wanted to catch Jim. However , although they started at the same time, Jim traveled at 80 km/h and Sarah traveled at 120 km/h . How much of a head start did Jim have if it took three hours for Sarah to catch him ?
Jim traveled at 80 km/h and Sarah traveled at 120 km/h
It took 3 hours for Sarah to catch Jim.
Let's find out how much distance both covered.
[tex]Jim\colon\; d=r\cdot t=80\cdot3=240\: km[/tex]So, Jim traveled 240 km
[tex]Sarah\colon\; d=r\cdot t=120\cdot3=360\: km[/tex]So, Sarah traveled 360 km
This means that Jim must have started 360 - 240 = 120 km ahead.
Therefore, Jim had a head start of 120 km
in the graph of y= 8x + 5, 8 is theof the line
the function is:
[tex]y=8x+5[/tex]where 8 is the slope of the function
Given a student has a dog, what is the probability that a student also has a cat?62.9%57.1%41.8%36.3%
The given problem is a conditional probability problem.
Probability that a student has a cat given that he/she has a dog is represented as:
[tex]Pr(C|D)=\frac{Pr(CnD)}{Pr(D)}[/tex][tex]\begin{gathered} \text{CnD}=16 \\ \text{Sample space=28+16+24}=68 \end{gathered}[/tex]Thus,
[tex]Pr(\text{CnD)}=\frac{16}{68}=0.2353[/tex][tex]\begin{gathered} Number\text{ of dogs only=28} \\ Pr(D)=\frac{28}{68}=0.4117 \end{gathered}[/tex]Therefore,
[tex]undefined[/tex]y = -x - 2 y + 2 = -x Graph each system. Tell whether the system hasA.no solutionB.one solutionC. infinitely many solutionsD. Cannot determine
To graph each equation in the system, you can give it x-values, plug into the equations, and get values for Y.
Since a single line passes through two points, just take two values of x for each equation. So, for the first you have for example
*If x = 3
[tex]\begin{gathered} y=-x-2 \\ y=-3-2 \\ y=-5 \\ \text{ So} \\ (3,-5) \end{gathered}[/tex]*If x = -4
[tex]\begin{gathered} y=-x-2 \\ y=-(-4)-2 \\ y=4-2 \\ y=2 \\ \text{ So,} \\ (-4,2) \end{gathered}[/tex]For the second equation you have for example
*If x = 1
[tex]\begin{gathered} y+2=-x \\ y+2=-1 \\ y+2-2=-1-2 \\ y=-3 \\ \text{ So,} \\ (1,-3) \end{gathered}[/tex]*if x = -1
[tex]\begin{gathered} y+2=-(-1) \\ y+2=1 \\ y+2-2=1-2 \\ y=-1 \\ \text{ So,} \\ (-1,-1) \end{gathered}[/tex]Now, graphing the equations you have
As you can see, the lines associated with this system of equations overlap, that is, they share infinite solution points.
Therfore, the correct answer is C. infinitely many solutions.
The doctor orders 3000 mL D5RL to run at 300 ml/hr. How long will this IV infusion run
we know that
the infusion run at 300 ml/hr
so
Applying proportion
Find out how long for 3,000 ml
Which inequality is true when the value of u is 1? O-- 10 > -1.5 u – 10 > 1.5 O –u – 10 – 1.5 O – 10 >1.5
First, let's solve each one of the inequalities for u
1) -u-10>-1.5
u<-10+1.5
u<-8.5 then this statement is not true for u=1
2) u-10>1.5
u>1.5+10
u>11.5 then this statement is not true for u=1
3) -u-10<1.5
u>-10-1.5
u>-11.5 which is true for u=1
4)-u-10>1.5
u<-11.5 which is not true for u=1
then the correct option will be -u-10<1.5If the fish tanks dimension are 60 by 15 by 34 and its is completely empty, what volume of water is needed to fill three fourths of the aquarium? Please help what would the volume if you only filled 3/4 of the tank
First let's find the volume of the fish tank. Given that the dimensions are 60 by 15 by 34, then:
[tex]V=(60)(15)(34)=30600[/tex]we have that the total volume of the fish tank is 30600 u³. But we only want to know how much is 3/4 of the total volume, then:
[tex](30600)(\frac{3}{4})=22950[/tex]therefore, to fill three fourths of the aquarium we will need 22950 u³ of water
The average daily high temperatura for the month of may in Ocala, Florida is approximated by the fuction f(n)= 0.2n + 80, where n is the day of the month. May has 31 days. The maximum daily high temperature ocurred on May 31 st. What was the msximum temperature?
n = day of the month
Therefore,
The maximum temperature for may 31st will be
[tex]\begin{gathered} f(n)=0.2n+80 \\ f(31)=0.2(31)+80 \\ f(31)=6.2+80=86.2\text{ } \end{gathered}[/tex]a railroad tracks can be determined using the following graph. Several different rosdways are in the same region as the railroad. Part B: A turnpikes route is determined by the equation y=1/3x^2. Prove algebraically how many intwrsections there will be between the railroad abd the turnpike,showing all necessary work
Given
[tex]y=\frac{1}{3x^2}[/tex]2x+3y=18
Find
Prove algebraically how many intwrsections there will be between the railroad
Explanation
The graph of 2x+3y=18 is as the picture
2x+3y=18
when x=0, 0+3y=18 => y=6 =>(0,6)
when y=0, 2x+0=18 => x=9 => (9,0)
The intersection between the railroad and the highway is 0 because the graph of the railroad and the graph of the highway are parallel, that means they have no intersection
(b)
Assume the railroad can be found using the equation y=3/2x+b
when x=0 => y=8
[tex]\begin{gathered} \frac{1}{3}x^2=\frac{3}{2}x+8 \\ 2x^2-9x-48=0 \\ D=9^2-4(2)(-48)=465 \\ =>D>0 \\ \frac{1}{3}x^2=\frac{3}{2}x+8 \end{gathered}[/tex]has two roots, and there are 2 intersections
Final Answer
(a) No intersection
(b) Two intersections
Find greatest common factor for each group,factor completely and find real roots
SOLUTION
Write out the polynomial given
The first group of the expresion is
[tex]\begin{gathered} 3x^3+4x^2 \\ \text{Then the GCE is } \\ x^2(\frac{3x^3}{x^2}+\frac{4x^2}{x^2}) \\ \text{GCE}=x^2 \end{gathered}[/tex]GCE is x²
For the second group, we have
[tex]\begin{gathered} 75x+100 \\ \text{GCE}=25(\frac{75x}{25}+\frac{100}{25}) \\ \text{GCE}=25 \end{gathered}[/tex]The GCE for the secod group is 25
To factorise completely, we have
[tex]\begin{gathered} 3x^3+4x^2+75x+100 \\ \\ x^2(\frac{3x^3}{x^2}+\frac{4x^2}{x^2})+25(\frac{75x}{25}+\frac{100}{25}) \end{gathered}[/tex]Then by simplification, we have
[tex]\begin{gathered} x^2(3x+4)+25(3x+4) \\ \text{Then, we factor completely to get} \\ (3x+4)(x^2+25) \end{gathered}[/tex]Then factors are (3x +4)(x²+ 25)
To find the real root, we equate each of the factors to zero, hence
[tex]\begin{gathered} (3x+4)(x^2+25)=0 \\ \text{Then} \\ 3x+4=0orx^2+25=0 \\ 3x=-40rx^2=-25 \\ \end{gathered}[/tex]Thus
[tex]\begin{gathered} \frac{3x}{3}=-\frac{4}{3} \\ x=-\frac{4}{3}\text{ is a real root } \\ or\text{ } \\ x^2=-25 \\ \text{take square root} \\ x=\pm_{}\sqrt[]{-25}\text{ not a real root} \end{gathered}[/tex]Therefore, since the root of -25 is a complex number,
The only real root is x = -4/3
-Determine whether each relation is a function. Explain your answerA. {(7,4),(6,3),(5,2)}B. {(15,0),(15,-2)}C. {(0,1),(2,1),(0,3)}
A. it is a fuction because for every number in the first position of the pairs it is one and only one second position number.
B. it's not a function because the number 15 has two different pairs
C. it's not a function because zero has two different pairs, 1 and 3.
Which of the geometric objects are scaled versions ofeach other?For the objects that are scaled versions of each otherfill out the table withFigure Type- Figure Labels (smallest to largest)-Figure Ratio (might be extended ratio)Ive filled out the circles for you as an example and because the ratio involves radials. Figure labels| figure ratioCircle- J,L,G | √2: √ 5 :3
I will do the squares for you
All squares are similar
E K and C are squares so they are scaled versions of each other
The length of the top side of E 2 is units
The length of the top side of K is 3 units
The length of the top side of C is 5 units
From smallest to largest is E, K, C
Figure ratio is 2,3,5
The rectangles are done in the same manner
D is a 2 by 4
F is a 2 by 6
I is a 3 by 6
D and I are similar
D,I the ratio is 2:3
Triangles work like rectangles
B is a 2 by 3
H is a 3 by 5
A is a 4 by 6
B and A are similar
B,A the ratio is 2:4
(5 points each for a and b) A bacteria colony starts with 20 bacteria andgrows continuously at a rate of 28% per hour.a. How long will it take for the colony to:i. Double its size?ii. Reach 500,000 bacteria?b. How many bacteria will there be in:i. 3 hours?ii. 3.5 days?
Given:
The initial population of bacteria, I = 20
Growth rate, r = 28%
Explanation:
a) To find: The time
i) Double its size
Using the formula,
[tex]F=I(1+r)^t,\text{ Where I denotes initial and F denotes Final size.}[/tex]On substitution we get,
[tex]\begin{gathered} 40=20(1+0.28)^t \\ 1.28^t=\frac{40}{20} \\ t=\log _{1.28}2 \\ t=2.81 \end{gathered}[/tex]Thus, the answer is 2.81 hours.
ii) To reach 500000 bacteria:
[tex]\begin{gathered} 500000=20(1+0.28)^t \\ 1.28^t=\frac{500000}{20} \\ t=\log _{1.28}(25000) \\ t=41.02\text{ hours} \end{gathered}[/tex]Thus, the answer is 41.02 hours.
b) To find the bacteria size:
i) In 3 hours,
[tex]\begin{gathered} F=20(1+0.28)^3 \\ =41.94304 \\ \approx42 \end{gathered}[/tex]Thus, the size of bacteria in 3 hours is 42.
i) In 3.5 days,
That is, 84 hours
[tex]\begin{gathered} F=20(1+0.28)^{84} \\ =20261306488.67 \\ \approx20261306489 \end{gathered}[/tex]Thus, the size of bacteria in 3.5 days is 20261306489.
Identify the statement that best describes the output for the following R command. filter (Fingers, SSLast != “NA”)a) A list of the values in SSLast that are “NA”.b) A list of the values in SSLast excluding the “NA” values.c) A list of all the values in SSLast, but not in Fingers.d) A list of all values in Fingers excluding the “NA” values.
solution
filter(Fingers, SSLast != "NA") includes only cases for which the variable SSLast is not equal to NA. Therefore
answer:
b) A list of the values in SSLast excluding the “NA” values.
In the accompanying diagram of parallelogramABCD, m_A = (2x + 10) and mZB = 3x. Find thenumber of degrees in m_B.D3x(2x + - 10)ABBYour answer
Answer
Angle B = 102°
Angle A = 78°
Explanation
The first thing to note in answering this is that for parallelograms,
- Opposite angles are equal to each other.
- Adjacent angles (angles close to each other) sum up to give 180°.
So, in the given parallelogram, we can see that the two angles given are adjacent angles. Hence,
2x + 10° + 3x = 180°
5x + 10° = 180°
5x = 180° - 10°
5x = 170°
Divide both sides by 5
(5x/5) = (170°/5)
x = 34°
So, we can solve for Angle B now
Angle B = 3x = 3 (34°) = 102°
Angle A = 2x + 10° = 2(34°) + 10° = 68° + 10° = 78°
Hope this Helps!!!
given the polygon below, if
The value of ∠Q = 130
A polygon is a flat or plane two-dimensional closed shape with straight sides. It doesn't have any curved edges. A polygon's sides are also known as its edges.
Given that in the polygon ∠T = ∠S and ∠S = 115
We have to find q
The formula to calculate the sum of the inner angles of an n sides polygon is
(n-2) x 180
= (5-2) x 180
= 3 x 180
= 540
Sum of inner angles of polygon = 540
Since T = S
In the polygon P = R = 90
Q = 540 – 2 x 115 – 2 x 90
Q = 540 – 230 – 180
Q = 540 – 410
Q = 130
Therefore the value of ∠Q = 130
To learn more about polygons visit
https://brainly.com/question/10441863
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Choose the linear equation that best fits the data on the graph
We have several points in a graph and want to know what of the options is the best fit.
Due the options has very diferent slopes for the line we can solve the problem by inspection, so:
• The slope of the line must be negative, because the points data show a negative slope. ,So, the options A and B are discarded.
,• The points show a big slope. For Option C, when x = 4, y = -7 that is near from data, for Option D, when x =4, y = -1 which is very far from data.
So the correct answer is the option C.
I need help with the Try it! section. i will flow a walk through if you can give one
Solution
For this case we need to remember that the general equation for a line is given by:
y= mx+ b
Where m represent the slope and b the intercept
And we can find the slope with this formuala:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]And we can use (50, 725) and (100,1325) and we have:
[tex]m=\frac{1325-725}{100-50}=12[/tex]And the intercept would be:
725 = 50*12 +b
b= 725 - 600
b= 125
and the equation would be given by:
y= 12 x + 125
And the y intercept represent the starting value of 125$ no matter the number of guests
What is the area of trapezoid KLMO?A) 224cm^2B) 112 cm^2C) 128 cm^2D) 96 cm^2
Given:
The length of the bases of the trapezoid
[tex]\begin{gathered} KL=a=12cm \\ \\ OM=b=16cm \end{gathered}[/tex]Height of the trapezoid:
[tex]LN=h=8cm[/tex]Required:
The area of trapezoid KLMO
Explanation:
The formula for area of trapezoid is given by
[tex]A(trapezoid)=\frac{a+b}{2}\times h[/tex]Substituting the given values in the above equation we get
[tex]\begin{gathered} A(trapezoid\text{ }KLMO)=\frac{a+b}{2}\times h \\ \\ A(trapezoid\text{ }KLMO)=\frac{12+16}{2}\times8=\frac{28}{2}\times8=14\times8=112cm^2 \end{gathered}[/tex]Final answer:
The area of trapezoid KLMO is 112 sq.cm