Answers
An important rule before subtracting fractions, you must first make them have a common denominator. The common denominator is the LCM (Least Common Multiple) of the two different denominators.
You may use this formula:
[tex]\frac{a}{b}=\frac{(a)(\frac{c}{b})}{c}[/tex]We get,
a.) 9/12 - 1/2 ; The LCM of 12 and 2 is 12.
[tex]\frac{9}{12}-\frac{1}{2}=\text{ }\frac{9}{12}-\frac{(1)(\frac{12}{2})}{12}=\text{ }\frac{9}{12}-\frac{(1)(6)}{12}=\frac{9}{12}-\frac{6}{12}[/tex][tex]\frac{9}{12}-\frac{6}{12}=\frac{3}{12}[/tex]Therefore, the difference of 9/12 - 1/2 is 3/12. You shade letter B.
Related Questions
CJ loves Girl Scout cookies. He eats 3 cookies per hour. After 5 hours, there are 24 cookies left in the box. Write an equation in slope intercept form. Determine how many hours it will take CJ to eat the entire box of cookies.
Answers
To solve this problem: y will represent the number of cookies, and x the number of hours.
To find the number of cookies that CJ eats per hour, we multiply 3 (since he eats 3 per hour) by x (the number of hours)
Since there we only 24 cookies left in the box, we will need to substract 3 by the number of hours that have passed, from 24 to find the number of cookies "y":
[tex]y=24-3(x-5)[/tex]This equation represents that the number of cookies "y" is equal to the 24 cookies that where left after 5 hours, and to that we substract 3 (which is the number of cookies per hour) by total number of hours that have passed since those 5 hours (x-5) because 5 hours that have already passed we substract them from x.
We need to simplify that equation to represent in slope-intercept form:
[tex]\begin{gathered} y=24-3x+15 \\ y=-3x+39 \end{gathered}[/tex]Now we need to determine the number of hours it would take to finish the cookies. So we are looking for the value of x, that makes y=0:
[tex]0=-3x+39[/tex]solving for the number of hours x:
[tex]\begin{gathered} -3x=-39 \\ x=-\frac{39}{(-3)} \\ x=13 \end{gathered}[/tex]It would take 13 hours for CJ to eat the entire box of cookies.
According to a pencil company's advertising campaign, 2 out of 3 students prefer the company's pencils to their competitor's pencils. A representative of the company went to a high school with 1479 students for alq & A session, and one of the students asked, "If this is true, in this high school, how many more students are there who prefer your pencils than students who prefer your competitor's pencils?" Help the company's representative come up with an answer. Assume that the company's claim is true.Part 1: How many of the students in the high school prefer the company's pencils? Part II: Out of 3 students, how many prefer the pencils of the company's competitor?Part III: How many of the students in the high school prefer the pencils of the company's competitor? Part IV: In the high school, how many more students are there who prefer the company's pencils than students who prefer their competitor's pencils?
Answers
Data:
2 of 3 students prefer the company's pencils to their competitor's pencils
high school with 1479 students
Part 1:You have 1479 students in total and know that 2 of 3 students prefer the company's pencils, then you multiply the toatl number of students by the factor 2/3
[tex]1479\cdot\frac{2}{3}=\frac{2958}{3}=986[/tex]Then, 986 students in the high school prefer the company's pencilsPart 2:Of 3 students 1 prefer the pencils of the company's competitorPart 3:Substract the number of students that prefer the company's pencils from the total of students:
[tex]1479-986=493[/tex]Then, 493 students in the high school prefer the pencils of the company's competitorPart 4:Substract the number of students who prefer the competitor's pencils fom the students that prefer the company's pencils:
[tex]986-493=493[/tex]Then, in the high school there are 493 students more that prefer the company's pencils than students who prefer their competitor's pencilsFor a standard normal distribution, find the z-value that goes with a left tail area=0.9931
Answers
The z-value that goes with a left tail Area= 0.9931 is 2.4 .
What is normal distribution?A probability distribution that is symmetric about the mean is the normal distribution, also known as the Gaussian distribution. Data close to the mean are more common than data far from the mean. The normal distribution is displayed as a "bell curve" on the chart.
What is left rear area?
The area under the curve to the left of x* in Figure 5.19, “Right and left tails of the distribution” is known as the left tail of the density curve for a continuous random variable X whose limit is x* (a).
According to the Z-value normal distribution table, its value is 2.4 .
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If P = (-3,5), find the imageof P under the following rotation.180° counterclockwise about the origin([?], []).Enter the number that belongs inthe green box.Enter
Answers
The rule for a 180° counterclockwise rotation is-
[tex](x,y)\rightarrow(-x,-y)\text{.}[/tex]So, we just have to change the sign of each coordinate.
[tex](-3,5)\rightarrow(3,-5)[/tex]Hence, the image is (3,-5).Apply zero product theorem to solve for x[tex]x ^{2} = 9[/tex]
Answers
Answer:
[tex]\begin{gathered} x_1=-3 \\ x_2=3 \end{gathered}[/tex]Step-by-step explanation:
To apply the zero product theorem, put all the terms on the left side to equal zero.
[tex]x^2-9=0[/tex]Factoring the binomial:
[tex]\begin{gathered} (x+3)(x-3)=0 \\ x_1+3=0 \\ x_1=-3 \\ \\ x_2-3=0 \\ x_2=3 \end{gathered}[/tex]A corporation distributes a 10% common stock dividend on 30000 shares issued when the market value of its common stock is $24 per share and its par value is $2 per share dollars per share on the distribution date a credit for $___ would be journalized.A. $30,000B. $6,000C. $72,000D. $66,000
Answers
A corporation distributes a 10% common stock dividend on 30,000 shares.
The market value is $24 per share.
The par value is $2 per share.
We have to find the credit that is journalized the moment the distribution is made.
They paid a total amount in dividends that is 10% of the par value of the stock times the number of stocks:
[tex]\begin{gathered} 10\%\cdot2\cdot30000 \\ 0.1\cdot2\cdot30000 \\ 6000 \end{gathered}[/tex]Answer: the credit is $6,000 [Option B]
8 with a exponent of 3 divided by 2
Answers
8³ ÷ 3
First we find the value of 8³;
8³ = 512
Then divide by 3
512/3 = 170.6
Weekly wages at a certain factory arenormally distributed with a mean of$400 and a standard deviation of $50.Find the probability that a workerselected at random makes between$500 and $550.
Answers
The Solution:
Step 1:
We shall state the formula for calculating Z-score.
[tex]\begin{gathered} Z=\frac{X-\mu}{\sigma} \\ \text{Where X}=5\text{00 ( for lower limit) and X=550 for upper limit.} \\ \mu=400 \\ \sigma=50 \end{gathered}[/tex]Step 2:
We shall substitute the above values in the formula.
[tex]\begin{gathered} \frac{500-400}{50}\leq P(Z)\leq\frac{550-400}{50} \\ \\ \frac{100}{50}\leq P(Z)\leq\frac{150}{50} \\ \\ 2\leq P(Z)\leq3 \end{gathered}[/tex]Step 3:
We shall read the respective probabilities from the Z score distribution tables.
From the Z-score tables,
P(3) = 99.9 %
P(2) = 97.7 %
Step 4:
The Conclusion:
The probability that a worker selected makes between $500 and $550 is obtained as below:
[tex]\text{Prob}(500\leq Z\leq550)=99.9-97.7\text{ = 2.2 \%}[/tex]Therefore, the required probability is 2.2 %
Alex surveyed 60 student about their Vera zoo animals and made the circle graph of the results shown below
Answers
we can use the cross multiplication
we know that the full angle of a circle is 360° so the total angle corresponds to 60 students
so, what is 72 degrees?
[tex]\begin{gathered} 360\longrightarrow60 \\ 72\longrightarrow x \end{gathered}[/tex]where x is the number of students than said giraffes
[tex]\begin{gathered} x=\frac{72\times60}{360} \\ \\ x=12 \end{gathered}[/tex]the students than said giraffes are 12
x + y =5 x + y = 6 one solution no solutions infinitely many solutions
Answers
Problem
x + y = 5
x + y = 6
method
A system has no solution if the equations are inconsistent, they are contradictory.
for example
2x + 3y = 10
2x + 3y = 12 has no solution.
Final answer
x + y = 5
x + y = 6
are inconsistent
hence, the equations has no solution
NO SOLUTION
Triangles ABE, ADE, and CBE are shown on the coordinate grid, and all the vertices have coordinates that are integers. Which statement is true?
Answers
To check if Triangles ABE, ADE, and CBE are congruent, let us compute for the distance of each line using the Distance Formula,
[tex]\text{ }d\text{ = }\sqrt[]{(x_2-x_1)^2\text{ + (}y_2-y_1)^2}[/tex]Where,
d = Distance
(x1, y1) = Coordinates of the first point
(x2, y2) = Coordinates of the second point
Let's compute the distance of the following lines:
Triangle ABE: Lines AB, AE, and BE
Triangle ADE: Lines AD, AE, and ED
Triangle CBE: Lines CE, CB, and BE
For Triangle ABE,
[tex]\text{ d}_{AB}\text{ = }\sqrt[]{(-1-(-4))^2+(3-(-1))^2}\text{ = }\sqrt[]{(-1+4)^2+(3+1)^2}[/tex][tex]\text{ d}_{AB}\text{ = }\sqrt[]{(3)^2+(4)^2}\text{ = }\sqrt[]{9+\text{ 16}}\text{ = }\sqrt[]{25}[/tex][tex]\text{ d}_{AB}\text{ = 5}[/tex][tex]\text{ d}_{AE}\text{ =}\sqrt[]{(1-\text{ }(-1))^2+(0\text{ - }(-4))^2}\text{ = }\sqrt[]{(1+1)^2+(0+4)^2}[/tex][tex]\text{ d}_{AE}\text{ = }\sqrt[]{(2)^2+(4)^2}\text{ = }\sqrt[]{4\text{ + 16}}[/tex][tex]\text{ d}_{AE}\text{ =}\sqrt[]{20}[/tex][tex]\text{ d}_{BE}\text{ = }\sqrt[]{(1\text{ - (}3))^2+(0\text{ - (-1)})^2}\text{ = }\sqrt[]{(1-3)^2+(0+1)^2}[/tex][tex]\text{ d}_{BE}=\text{ }\sqrt[]{(-2)^2+(1)^2}\text{ = }\sqrt[]{4\text{ + 1}}[/tex][tex]\text{ d}_{BE}\text{ = }\sqrt[]{5}[/tex]For Triangle ADE, let's compute for the distance of line AD and ED since we already got the distance of line AE.
[tex]\text{ d}_{AD}\text{ = }\sqrt[]{(-1-(-1))^2+\text{ (}1\text{ - }(-4))^2}\text{ = }\sqrt[]{(-1+1)^2+(1+4)^2}[/tex][tex]\text{ d}_{AD}\text{ = }\sqrt[]{(0)^2+(5)^2}\text{ = }\sqrt[]{25}[/tex][tex]\text{ d}_{AD}\text{ = 5}[/tex][tex]\text{ d}_{ED}=\text{ }\sqrt[]{(-1\text{ - (}1))^2+(1-0)^2}\text{ = }\sqrt[]{(-1-1)^2+(1)^2}[/tex][tex]\text{ d}_{ED}\text{ = }\sqrt[]{(-2)^2_{}+(1)^2}\text{ = }\sqrt[]{4\text{ + 1}}[/tex][tex]\text{ d}_{ED}\text{ = }\sqrt[]{5}[/tex]For Triangle CBE, let's compute for the distance of line CE and CB since we already got the distance of line BE.
[tex]\text{ d}_{CE}\text{ = }\sqrt[]{(3-\text{ }1)^2+(4-0)^2}\text{ = }\sqrt[]{(2)^2+(4)^2}[/tex][tex]\text{ d}_{CE}\text{ = }\sqrt[]{4+16}\text{ = }\sqrt[]{20}[/tex][tex]\text{ d}_{CE}\text{ = }\sqrt[]{20}[/tex][tex]\text{ d}_{CB}\text{ = }\sqrt[]{(3-3)^2+(4\text{ - (}-1))^2}\text{ =}\sqrt[]{(0)^2+(4+1)^2}[/tex][tex]\text{ d}_{CB}\text{ =}\sqrt[]{(5)^2}\text{ = }\sqrt[]{25}[/tex][tex]\text{ d}_{CB}\text{ = 5}[/tex]In summary,
Triangle ABE:
[tex]AB=\text{ 5, AE = }\sqrt[]{20}\text{ and BE = }\sqrt[]{5}[/tex]Triangle ADE:
[tex]\text{ AD = 5, AE = }\sqrt[]{20}\text{ and ED = }\sqrt[]{5}[/tex]Triangle CBE: CE, CB, and BE
[tex]\text{ CB = 5, CE = }\sqrt[]{20}\text{ and BE = }\sqrt[]{5}[/tex]The sides of the three triangles shown in the grid are congruent based on the SSS Rule of Triangle.
Thus, the statement that meets our evaluation is:
D. Triangle ABE, ADE and CBE are all congruent.
Solve the equation for all real solutions. 9z^2-30z+26=1
Answers
Weare given the following quadratic equation, and asked to find all its real solutions:
9 z^2 - 30 z + 26 = 1
we subtract "1" from both sides in order to be able to use the quadratic formula if needed:
9 z^2 - 30 z + 26 - 1 = 0
9 z^2 - 30 z + 25 = 0
we notice that the first term is a perfect square:
9 z^2 = (3 z)^2
and that the last term is also a perfect square:
25 = 5^2
then we suspect that we are in the presence of the perfect square of a binomial of the form:
(3 z - 5)^2 = (3z)^2 - 2 * 15 z + 5^2 = 9 z^2 - 30 z + 25
which corroborates the factorization of the trinomial we had.
Then we have:
(3 z - 5)^2 = 0
and the only way such square gives zero, is if the binomial (3 z - 5) is zero itself, which means:
3 z - 5 = 0 then 3 z = 5 and solving for z: z = 5/ 3
Then the only real solution for this equation is the value:
z = 5/3
this is factor by grouping. did I do 1a right and how do I continue on 1b
Answers
We will have the following:
*First: ( f ° g) (x):
[tex](f\circ g)(x)=\frac{(\frac{1}{x})+1}{(\frac{1}{x})-2}\Rightarrow(f\circ g)(x)=\frac{(\frac{1+x}{x})}{(\frac{1-2x}{x})}[/tex][tex]\Rightarrow(f\circ g)(x)=\frac{(1+x)(x)}{(x)(1-2x)}\Rightarrow(f\circ g)(x)=\frac{1+x}{1-2x}[/tex]Domain:
[tex](-\infty,\frac{1}{2})\cup(\frac{1}{2},\infty)[/tex]*Second: (f ° f) (x):
[tex](f\circ f)(x)=\frac{(\frac{x+1}{x-2})+1}{(\frac{x+1}{x-2})-2}\Rightarrow(f\circ f)(x)=\frac{(\frac{(x+1)+(x-2)}{x-2})}{(\frac{(x+1)-2(x-2)}{x-2})}[/tex][tex]\Rightarrow(f\circ f)(x)=\frac{(\frac{2x-1}{x-2})}{(\frac{-x+5}{x-2})}\Rightarrow(f\circ f)(x)=\frac{(2x-1)(x-2)}{(x-2)(-x+5)}[/tex][tex]\Rightarrow(f\circ f)(x)=\frac{2x-1}{-x+5}[/tex]Domain:
[tex](-\infty,5)\cup(5,\infty)[/tex]In Mr. Johnson’s third and fourth period classes, 30% of the students scored a 95% or higher on a quiz.Let be the total number of students in Mr. Johnson’s classes. Answer the following questions, and showyour work to support your answer.If 15 students scored a 95% or higher, write an equation involving that relates the number ofstudents who scored a 95% or higher to the total number of students in Mr. Johnson’s third andfourth period classes. Of the students who scored below 95% 40% of them are girls. How many boys scored below 95%?
Answers
Total number of students = scored below 95% + scored above 95% (I)
______________________________
Students scored below 95%
Scored below 95%* 0.40 = girls
Boys = total scored below 95% (100% - 40%)
Boys = total scored below 95% (60%)
Boys = total scored below 95% (0.6) (II)
__________________________________________
Can you see the updates?
_______________________________
30% of the students scored a 95% or higher on a quiz and 15 students scored a 95% or higher
Total number of students* 30 = 15
30%*n = 15
n= 15/ 0.3
n= 50
_____________________________
Replacing in (I)
Total number of students = scored below 95% + scored above 95%
50 = 15 + 35
Replacing in (II)
Boys = 15 (0.60) = 9
__________________________________________
Answer
9 of the students who scored below 95% are boys
Carbon-14 is used for archeological carbon dating. Its half-life is 5730 years. How much of a 50-gram sample of Carbon-14 will be left in 1000 years?
Answers
Given:
The half-life of carbon-14 is 5730 years.
The initial amount of carbon is I = 50 grams.
Explanation:
To find the final amount of carbon after 1000 years.
The fundamental decay equation is,
[tex]\begin{gathered} F=Ie^{-\lambda t} \\ \text{Where, }\lambda=\frac{\ln 2}{t_{\frac{1}{2}}} \end{gathered}[/tex]Let us find the radioactive constant first.
[tex]\begin{gathered} \lambda=\frac{\ln 2}{5730} \\ \lambda=0.00012096809 \end{gathered}[/tex]Then, the final amount of the corban-14 is,
[tex]\begin{gathered} F=50e^{-0.000121(1000)}^{} \\ =44.30g \end{gathered}[/tex]Hence, the amount of a 50-gram sample of Carbon-14 will be left in 1000 years is 44.30 g.
? Question The table shows certain values of a fourth-degree polynomial function with no repeated factors. -12 -10 -6 -4 2 4 8 10 12 у 280 81 -14 0 0 -24 0 126 400 The function must have a zero between the x-values of -12 and -10 Between the x-values of 2 and 8, the graph of the function should be drawn the x- The function must be positive for all x-values between Submit
Answers
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Answer: (-10 and 06), below, (-4, 2)
Step-by-step explanation:
From Edmentum.
The Knitting Club members are preparing identical welcome kits for new members. The Knitting Club has 45 spools of yarn and 75 knitting needles. What is the greatest number of identical kits they can prepare using all of the yarn and knitting needles?
Answers
Common factors of 45 : 1,3,5,9,15,45
Common factors of 75 : 1,3,5,15,25,75
Common factors: 1,3,5,15
GReatest common factor = 15
15 identical kits
Multiples of 36 and the square root of 49
Answers
In this case, we'll have to carry out several steps to find the solution.
Step 01:
Data:
36
√49
Step 02:
multiples:
36:
The multiples of thirty-six are the numbers that contain it a whole number of times.
36, 72, 108, 144, 180, ...
√49:
[tex]\sqrt[]{49}=7[/tex]The multiples of seven are the numbers that contain it a whole number of times.
7, 14, 21, 28, 35, 42, ...
That is the full solution.
the beginning of the question is "if the slope of..." please help
Answers
we know that
If two lines are parallel, then their slopes are equal
In this problem
mAB=mCD
substitute the given values
3/4=(x-2)/12
solve for x
multiply by 12 both sides
(x-2)=12(3/4)
x-2=9
x=9=9+2
x=11what is 39 ÷ (2+ 1) - 2 × (4 + 1)
Answers
The given expression is
[tex]36\colon(2+1)-2\times(4+1)[/tex]First, we solve the additions inside the parenthesis
[tex]36\colon3-2\times5[/tex]Then, we solve the produce and the division
[tex]12-10[/tex]At last, we subtract
[tex]12-10=2[/tex]Hence, the answer is 2.state the solution for the quadratic equation depicted in the graph.
Answers
For this problem, we were provided with the graph of a quadratic equation, and we need to determine the solutions for this graph.
The solutions of a quadratic equation are the values of "x" that make the expression equal to "0". Therefore, we need to look at the graph for the values at which the graph crosses "y=0".
We have two points for this problem. The first one is approximately -5, and the second is 6.
URGENT!! ILL GIVE
BRAINLIEST!!!! AND 100 POINTS!!!!!
Answers
Answer:
I don't know the answer but I want to say something ...you can't just go around writing HELP!!! ILL GIVE 100 POINTS when your question only gives 5!!! it's just deceptive, if you want someone's help at least be honest! thank you for your time
Which is the closest to the area of the triangle in square centimeters?
Answers
Option c ) 40 is the closest to the area of the triangle in square centimeters .
Formula for Area of a right-angled triangle :
Area of a right-angled triangle = [ ( 1 / 2 ) * base * height ]
According to question ,
base = 10.1 cm
height = 8.2 cm
So , Area of triangle = [ ( 1 / 2 ) * 10.1 * 8.2 ]
= 41.41 [tex]cm^{2}[/tex]
This is closest to option c ) 40 .
Hence , option c ) 40 is the closest to the area of the triangle in square centimeters .
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A student has 5 pairs of plants, 2 shirts and 7 necklaces. He chooses one shirt, one pair of pants, and one necklace. How many different outfits could he make?
Answers
Answer:
70
Explanation:
By the rule of multiplication, we can calculate the number of different outfits as
5 x 2 x 7 = 70
pants shirts necklaces
Therefore, there are 70 different outfits
Using the compound interest formula, determine the total amount paid back and the monthly payment. Buying a $6000 used sedan taken out with $500 paid up front and the rest borrowed at 8.3%annual interest compounded daily (365 days per year) over 2 years.
Answers
The final value of an investment or loan with compound interest is given by:
[tex]FV=P(1+\frac{r}{m})^{m\cdot t}[/tex]Where P is the initial value (principal or loan), r is the annual interest rate, t is the duration of the investment/loan, and m is the number of compounding periods per year.
The following values are given in the problem:
P = $6000 - $500 = $5500
r = 8.3% = 0.083
t = 2 years
m = 365
Applying the formula:
[tex]FV=5500(1+\frac{0.083}{365})^{365\cdot2}[/tex]Calculating:
[tex]FV=5500(1+0.0002273926)^{730}[/tex]FV = $6493.03
The total amount paid back is $6493.03
This is equivalent to an approximate monthly payment of:
[tex]R=\frac{$ 6493.03 $}{24}=270.54[/tex]The monthly payment is approximately $270.54
The center of a circle and a point on the circle are given. Writecenter: (3,2), point on the circle: (4,3)
Answers
Given:
The center of the circle is the point ( 3, 2 )
And the point on the circle is ( 4, 3 )
To write the equation of the circle, we need to find the radius
The radius = the distance between the center and the point on the circle
so, the radius is the distance between ( 3, 2) and ( 4, 3)
So,
[tex]\begin{gathered} r=d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2} \\ r=\sqrt[]{(4-3)^2+(3-2)^2^{}}=\sqrt[]{1+1}=\sqrt[]{2} \end{gathered}[/tex]The general equation of the circle is:
[tex](x-h)^2+(y-k)^2=r^2[/tex]Where (h, k) is the coordinates of the center of the circle, r is the radius
So,
[tex]\begin{gathered} (h,k)=(3,2) \\ r=\sqrt[]{2} \end{gathered}[/tex]so, the equation of the circle will be:
[tex](x-3)^2+(y-2)^2=2[/tex]which equation has a solution of x = 4
Answers
Answer
Option B is correct.
Only the equation, 3x + 9 = 21, has a solution of x = 4.
Explanation
We are told to pick the equation(s) with x = 4 as a solution from the equations,
5x - 8 = 44
3x + 9 = 21
4x = 24
x - 10 = -8
The step to solving this is to insert x = 4 and check if that is consistent with the given equation.
Option A
5x - 8 = 44
If x = 4
5(4) - 8 = 44
20 - 8 = 44
12 ≠ 44
Hence, this is not an answer
Option B
3x + 9 = 21
If x = 4
3(4) + 9 = 21
12 + 9 = 21
21 = 21
Hence, this is an answer for this question.
Option C
4x = 24
If x = 4
4(4) = 24
16 ≠ 24
Hence, this is not an answer to this question.
Option D
x - 10 = - 8
If x = 4
4 - 10 = -8
-6 ≠ -8
This is also not an answer to this question.
Hope this Helps!!!
Hello, how are you? I would like you to help me solve this exercise, please!
Answers
Given
Formula for the length of an arc
[tex]\begin{gathered} Measureof\text{Arc of a circle=}\frac{\theta}{360}\times2\pi r \\ \\ \end{gathered}[/tex]Parameters;
[tex]\begin{gathered} \theta=?\text{ , r=1m} \\ \text{measure of arc =}\frac{\pi}{9} \end{gathered}[/tex]We can substitute into the formula
[tex]\begin{gathered} \frac{\pi}{9}=\frac{\theta}{360}\times2\times\pi\times1 \\ \\ \frac{\pi}{9}=\frac{2\theta\pi}{360} \\ \text{cross multiply} \\ 18\theta\pi=360\pi \\ divide\text{ both sides by 18}\pi \\ \frac{18\theta\pi}{18\pi}=\frac{360\pi}{18\pi} \\ \theta=20^0 \end{gathered}[/tex]Now, change to radian
[tex]\frac{\pi}{180}\times20^0=\frac{20^0\pi}{180^0}=\frac{\pi}{9}[/tex]The final answer
[tex]\frac{\pi}{9}[/tex]Sue, who is 5 feet tall, is standing at Point D in the drawing. The tip of her head is a point E. a tree in the yard is at point B with the top of the tree at point C. Sue stand so her shadow meets at the end of the trees shadow at point a Which triangles similar?How do you know?Find the height of the tree (This distance from B to C).
Answers
Which triangles are similar?
The triangle AED and the triangle ABC is similar.
How do you know?
Because all the angles are equal, the triangle AED and ABC have the same angle values, then they're similar.
Find the height of the tree (This distance from B to C)
We can use the relation of the similar triangle to find BC, we can write the equation
[tex]\frac{AB}{AD}=\frac{BC}{ED}[/tex]The only unknown value here is BC, then
[tex]\frac{24+8}{8}=\frac{\text{BC}}{5}[/tex]Now we solve it for BC!
[tex]\begin{gathered} \frac{32}{8}=\frac{BC}{5} \\ \\ 4=\frac{BC}{5} \\ \\ BC=4\cdot5 \\ \\ BC=20\text{ ft} \end{gathered}[/tex]Hence, the height of the tree is 20 ft
The figure below shows the graph of f’ , the derivative of the function f, on the closed interval from x = -2 to x = 6. The graph of the derivative has horizontal tangentlines at x = 2 and x = 4.
Answers
Solution
- The points of inflection of f(x) in a graph of f'(x) is gotten by just finding the points where the graph moves from increasing to decreasing, and also from decreasing to increasing.
- Thus, we have
- The points where the graph changes from increasing to decreasing is at point (2, 0) and the point where the graph moves from decreasing to increasing is (4, -2.5)
- Thus, the inflection points of the graph of f are at (2, 0), and (4, -2.5)