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3 years ago

My answers are wrong plz help

Mathematics

2 answers:

Brilliant_brown [7]3 years ago

8 0

Your first answer is wrong.

The area should be:

a= 65

b= 109

The second one is probably C.)

lapo4ka [179]3 years ago

5 0

I think a would be 115 degrees. B would be 71 degrees. I think the second question would be c.

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For expressions a-e select yes or no to indicate whether each expression is equivalent to 3(x + 2).

Anuta_ua [19.1K]

Answer:

A. No

B. Yes

C. No

D. Yes

E. Yes

Step-by-step explanation:

3(x + 2) = 3x + 6

A. 3x + 2 is not 3x + 6

B. 3(2 + x) = 6 + 3x is equal to 3x + 6

C. 3x + 2x is not the same as 3x + 6

D. x + 2x + 2 + 4 = 3x + 6 is equal to 3x + 6

E. x + x + x + 1 + 1 + 1 + 1 + 1 + 1 = 3x + 6 is the same as 3x + 6

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3 years ago

Cylinders A and B are similar solids. The base of cylinder A has a circumference of 47 units. The base of cylinder B

egoroff_w [7]

Answer:

$\frac{3}{2}$ ( the third choice)

Step-by-step explanation:

Let $r_a$ be the radius of cylinder A and let $r_b$ be the radius of cylinder B. We assume the heights of both cylinders are the same

Base circumference of cylinder A = $2\pi r_a = 4\pi == > r_a = 2$

Base area of cylinder B = $\pi (r_b)^2 = 9\pi == > r_b = \sqrt{9} = 3$

In order to have $r_a$ (2) have the same value as $r_b$ (3) we must multiply by a factor of $\frac{3}{2}$

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2 years ago

-6 as a whole number

ValentinkaMS [17]

The Answer Would be 6

4 0

3 years ago

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Writing g for the acceleration due to gravity, the period,T, of a pendulum of length l is given by

ivanzaharov [21]

Step-by-step explanation:

$T=2\pi\sqrt{\frac{l}{g}}$

$T+\Delta T=2\pi\sqrt{\frac{(l+\Delta l)}{g}} =2\pi\sqrt{\frac{l}{g}}\sqrt{1+\frac{\Delta l}{l}}=2\pi\sqrt{\frac{l}{g}}(1+\frac{1}{2}\frac{\Delta l}{l}+0(\frac{\Delta l}{l}))=T(1+\frac{1}{2}\frac{\Delta l}{l}+0(\frac{\Delta l}{l}))$

Here, the Taylor approximation for a square root was applied, and O(x) stands for all negligible terms of Taylor's sum with respect to variable x.

So, $\Delta T=T\frac{1}{2}\frac{\Delta l}{l}$

b. For an increase of 2%, that is:

$\frac{\Delta l}{l}=0.02$

$\frac{\Delta T}{T}=\frac{1}{2}0.02=0.01=1\%$

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4 years ago

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jonny [76]

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