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3 years ago

11

how would i find the surface area of a cone in terms of pi, if the radius is 14cm & the height is 15cm ? what would be the a

nswer ?

1 answer:

RUDIKE [14]3 years ago

6 0

Answer:

196 * pi * square root of 421

Step-by-step explanation:

First, we would need to know the cone surface area formula.

πr (r+ square root of h^2+r^2

Let's plug in the values now!

We have

14π (14 + square root of 421) Open it up and we have

196 * pi * square root of 421

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!!!! help please !!!!

bogdanovich [222]

I think the answer might be A or D but most likely it's D

4 0

3 years ago

3+6×5+4÷2-7 using the same expression,add parentheses so that the value of the expression is 38

solmaris [256]

Answer:

3+6×(5+4÷2)-7

Step-by-step explanation:

To solve the expression use order of operations.

Right now the expression solves to:

3+6×5+4÷2-7 6* 5 = 30

3 + 30 + 4÷2-7 4 ÷ 2 = 2

3 + 30 + 2 - 7 Add and subtract left to right.

33 + 2 - 7

35 - 7

28

To make it solve to 38, add parenthesis:

3+6×(5+4÷2)-7 (5+4÷2) = 7

3+6×(7)-7 6*7 = 42

3 + 42 - 7 Add and subtract from left to right

45 - 7

38

7 0

3 years ago

find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6

taurus [48]

Answer:

$f(x) = \frac{6}{x^2} -8$ or $f(x) = -\frac{6}{x^2} + 4$

Step-by-step explanation:

Given

$(x,y) = (1,-2)$ --- Point

$\int\limits^6_2 {(1 + 144x^{-6})} \, dx$

The arc length of a function on interval [a,b]: $\int\limits^b_a {(1 + f'(x^2))} \, dx$

By comparison:

$f'(x)^2 = 144x^{-6}$

$f'(x)^2 = \frac{144}{x^6}$

Take square root of both sides

$f'(x) =\± \sqrt{\frac{144}{x^6}}$

$f'(x) = \±\frac{12}{x^3}$

Split:

$f'(x) = \frac{12}{x^3}$ or $f'(x) = -\frac{12}{x^3}$

To solve fo f(x), we make use of:

$f(x) = \int {f'(x) } \, dx$

For: $f'(x) = \frac{12}{x^3}$

$f(x) = \int {\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = \frac{12}{2x^2} + c$

$f(x) = \frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = \frac{6}{1^2} + c$

$-2 = \frac{6}{1} + c$

$-2 = 6 + c$

Make c the subject

$c = -2-6$

$c = -8$

$f(x) = \frac{6}{x^2} + c$ becomes

$f(x) = \frac{6}{x^2} -8$

For: $f'(x) = -\frac{12}{x^3}$

$f(x) = \int {-\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = -\frac{12}{2x^2} + c$

$f(x) = -\frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = -\frac{6}{1^2} + c$

$-2 = -\frac{6}{1} + c$

$-2 = -6 + c$

Make c the subject

$c = -2+6$

$c = 4$

$f(x) = -\frac{6}{x^2} + c$ becomes

$f(x) = -\frac{6}{x^2} + 4$

3 0

3 years ago

1.) Are the two expressions

Gnom [1K]

Answer:

yes

Step-by-step explanation:

if you distribute the 9 into the parenthesis you get 9y + 45. Therefore, both equations are equivalent seeing that they both equal to 9y + 45.

8 0

3 years ago

Read 2 more answers

Two hundred ten million, sixty-four thousand, fifty ¿in numbers?

marishachu [46]

210,64,000,50 is the anwser

7 0

3 years ago

Read 2 more answers