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2 years ago

Please help me with this

Mathematics

1 answer:

kherson [118]2 years ago

6 0

Answer:A or C

Step-by-step explanation: i guessed plus 19 hours ago

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The perimeter of the rectangle below is 132 units. Find the length of side AB.

melisa1 [442]

Answer:

AB = 35

Step-by-step explanation:

For the value of variable z....;

;Perimetre = 132

; 132 = 2(4z + 3) + 2(5z)

; 132 = 8z + 6 + 10z

; 132 - 6 = 8z + 10z

; 126 = 18z

;the value of...z = 7

Therefore the value of length AB;

AB = 5z...then substitute the value of z into the length of AB

; AB = 5(7)

;AB = 35

5 0

3 years ago

One day a store sold 33 sweatshirts, White ones cost $11.95 and yellow ones cost $11.50. In all, $386.70 worth of sweatshirts we

Contact [7]

16 whites $191.20 and 17 yellow $195.50 total $386.70

6 0

4 years ago

Which expression have the same value as the product of 2.7 and 4

Maurinko [17]

Answer:

Step-by-step explanation:

2.7*4=10.80

27*0.4=10.80 we multiplied 27 by 10 but divide 4 by 10

0.27*40=10.80

3 0

3 years ago

\lim _{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)

Vinvika [58]

$\displaystyle \lim_{x\to 0}\left(\frac{2x\ln \left(1+3x\right)+\sin \left(x\right)\tan \left(3x\right)-2x^3}{1-\cos \left(3x\right)}\right)$

Both the numerator and denominator approach 0, so this is a candidate for applying L'Hopital's rule. Doing so gives

$\displaystyle \lim_{x\to 0}\left(2\ln(1+3x)+\dfrac{6x}{1+3x}+\cos(x)\tan(3x)+3\sin(x)\sec^2(x)-6x^2}{3\sin(3x)}\right)$

This again gives an indeterminate form 0/0, but no need to use L'Hopital's rule again just yet. Split up the limit as

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} + \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} \\\\ + \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} + \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} \\\\ - \lim_{x\to0}\frac{6x^2}{3\sin(3x)}$

Now recall two well-known limits:

$\displaystyle \lim_{x\to0}\frac{\sin(ax)}{ax}=1\text{ if }a\neq0 \\\\ \lim_{x\to0}\frac{\ln(1+ax)}{ax}=1\text{ if }a\neq0$

Compute each remaining limit:

$\displaystyle \lim_{x\to0}\frac{2\ln(1+3x)}{3\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{\ln(1+3x)}{3x} \times \lim_{x\to0}\frac{3x}{\sin(3x)} = \frac23$

$\displaystyle \lim_{x\to0}\frac{6x}{3(1+3x)\sin(3x)} = \frac23 \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\frac{1}{1+3x} = \frac23$

$\displaystyle \lim_{x\to0}\frac{\cos(x)\tan(3x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\cos(x)}{\cos(3x)} = \frac13$

$\displaystyle \lim_{x\to0}\frac{3\sin(x)\sec^2(x)}{3\sin(3x)} = \frac13 \times \lim_{x\to0}\frac{\sin(x)}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}\sec^2(x) = \frac13$

$\displaystyle \lim_{x\to0}\frac{6x^2}{3\sin(3x)} = \frac23 \times \lim_{x\to0}x \times \lim_{x\to0}\frac{3x}{\sin(3x)} \times \lim_{x\to0}x = 0$

So, the original limit has a value of

2/3 + 2/3 + 1/3 + 1/3 - 0 = 2

6 0

3 years ago

A high school principal reports that the average sat math score for last year’s graduating class is 520. a sample is taken of th

Murljashka [212]

Based on the average SAT Math score that the graduating classes got, the comparison is that this year's graduating class did better than last year's graduating class.

<h3>How can both classes be compared?</h3><h3 />

The average score is used to show the general trend of scores that students were able to attain.

The average score of 530 that this year's graduating class attained is higher than the average of the previous graduating year which is 520.

This means that on average, this year's graduating class did better than that of last year because their general trend was higher.

Find out more on interpreting averages at brainly.com/question/1136789

#SPJ1

3 0

2 years ago