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3 years ago

I need help with this one to / surface area :)

Mathematics

2 answers:

nadezda [96]3 years ago

7 0

Step-by-step explanation:

the surface area (base area plus side wall) of a cone is, when we have the radius and the length of the slanted side wall :

SA = pi×r×(r + l)

in our case here

r = half of the diameter = 12/2 = 6 m

l = 13.4 m

so, we have

SA = pi×6×(6 + 13.4) = pi×6×19.4 = 365.6813849... m²

zzz [600]3 years ago

5 0

Answer:

389.9 to 1 decimal place

If you need another explanation comment on this one

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jill mowed lawns for 3 hours and earned $7.20 per hour.Then she washed windows for 4 hours and earned $6.75 per hour. what were

Aleksandr-060686 [28]

Actually, she worked for seven hours, so I'll solve for that.

7.20 x 3 = 21.6

6.75 x 4 = 27

21.6 + 27 = 48.6

48.6 /2 = 24.3

Her average should be $24.3 If she did work six hours, just multiply 6.75 x 3 and do what I did.

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3 years ago

What property is shown by the equation?

galben [10]

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3 years ago

Help for brainliest

Anarel [89]

Answer:

Rotation of 180 degrees, either direction

Dilation, making it smaller by some factor(we aren't given measurements so we cannot be exact)

Then move the figure to the right and then move the figure up

Step-by-step explanation:

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3 years ago

Find maclaurin series

Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

$\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

$\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}$

The first 3 terms of the series are

$\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}$

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

$\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}$

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

$\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}$

7 0

2 years ago

Given f(x) = 10 - 16/x, find all c in the interval [2,8] that satisfy the mean value theorem

Katen [24]

F(x)=10-16/x
f'(x)=16/x^2
f'(c)=16/c^2
f(8)=10-16/8=10-2=8
f(2)=10-16/2=10-8=2
$f'(c)= \frac{f(b)-f(a)}{b-a} in (2,8),
 \frac{16}{c^2} = \frac{f(8)-f(2)}{8-2} ,
 \frac{16}{c^2} = \frac{8-2}{8-2}
$
4∈[2,8] is the required point.

8 0

3 years ago

Read 2 more answers