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ANTONII [103]
2 years ago
7

Can anyone help me I really need help on this what I will give a brain for this

Mathematics
1 answer:
Hatshy [7]2 years ago
7 0

well since the formula is pi times radius squared that means it's 4 pi times 4 pi since the radius is squared so it's 16 pi

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PLEASE ANSWER ASAP!!! ANSWER MUST MUST MUST MMMUUUUSSSSSTTTTTT BE WITH AN EXPLANATION IN ORDER TO GET POINTS AND THE BRAINLIEST
Gelneren [198K]
V=LWH
SA=2(LW+LH+WH)
V=240
first check that they are 240 volume
multiply numbers they should equal 240

the first one (A) can be eliminated since the dimentions are the same and therefor the surface areas are the same

2nd one check so far

3rd one (C) 6*4*15=360 which is wrong so eliminate this one

4th (d) 12*6*5=360 which is not 240 so eliminate


answer is B
3 0
3 years ago
use the formulas for lowering powers to rewrite the expression in terms of the first power of cosine cos^4
Firdavs [7]

The expression cos⁴ θ in terms of the first power of cosine is <u>[ 3 + 2cos 2θ + cos 4θ]/8.</u>

The power-reducing formula, for cosine, is,

cos² θ = (1/2)[1 + cos 2θ].

In the question, we are asked to use the formulas for lowering powers to rewrite the expression in terms of the first power of cosine cos⁴ θ.

We can do it as follows:

cos⁴ θ

= (cos² θ)²

= {(1/2)[1 + cos 2θ]}²

= (1/4)[1 + cos 2θ]²

= (1/4)(1 + 2cos 2θ + cos² 2θ] {Using (a + b)² = a² + 2ab + b²}

= 1/4 + (1/2)cos 2θ + (1/4)(cos ² 2θ)

= 1/4 + (1/2)cos 2θ + (1/4)(1/2)[1 + cos 4θ]

= 1/4 + cos 2θ/4 + 1/8 + cos 4θ/8

= 3/8 + cos 2θ/4 + cos 4θ/8

= [ 3 + 2cos 2θ + cos 4θ]/8.

Thus, the expression cos⁴ θ in terms of the first power of cosine is <u>[ 3 + 2cos 2θ + cos 4θ]/8</u>.

Learn more about reducing trigonometric powers at

brainly.com/question/15202536

#SPJ4

5 0
1 year ago
Who know how to do this ?
nadya68 [22]
It seems to me this just wants you to add the 3 equations.
2m + 2n +5p


3 0
3 years ago
If YZ= 55 millimeters, what would be the length of QR? Millimeters
rusak2 [61]

Answer:

<h2>30 mm</h2>

Step-by-step explanation:

<h3>Hope it help </h3>
6 0
2 years ago
What are the vertical and horizontal asymptotes for the function f(x)=3^2/x^-4
Zarrin [17]

The vertical asymptotes x = - 2, 2 and Horizontal asymptotes y = 3

Option A is the correct answer.

<h3>What are Function?</h3>

Function are mathematical statement which links an independent variable to dependent variable.

It always comes with a defined domain and range.

The equation mentioned is incorrect in the question w.r.t. to the solutions mentioned, The right equation is

\rm f(x) = \dfrac{3 x^2}{x^{2}-4}

This function can also be written as

\rm f(x) = \dfrac{3 x^2}{x^{2}-2^2}\\\\\rm f(x) = \dfrac{3 x^2}{(x+2)(x-2)}

Vertical asymptotes are defined when the denominator of a rational function tends to zero.

To find the equation/s let the denominator equal zero.

Denominator tends to zero for x = -2 and x = +2

Horizontal Asymptote is when the function f(x) is tending to zero.

for x = +∞ and x = - ∞

Let the rational function f(x)= \dfrac{mx^a}{nx^b}  where  and a and b are degree of numerator and denominator.

If a < b , then the y axis y = 0 is a horizontal asymptote.

If b = a then horizontal asymptote is the line y = p/q

Here the value is b = 2 and a = 2

Since b = a the horizontal asymptote is the line

y = m/n  where m = 3 and n = 1

Therefore y = 3

So , vertical asymptotes x = - 2, 2

Horizontal asymptotes y = 3

Option A is the correct answer.

To know more about Function

brainly.com/question/12431044

#SPJ1

8 0
1 year ago
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