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3 years ago

What is the area of the composite figure?

Mathematics

2 answers:

earnstyle [38]3 years ago

4 0

Answer:

The answer is a

Step-by-step explanation:

jek_recluse [69]3 years ago

3 0

i think its b correct me if im wrong

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I’ll give the Brainliest to who answers this question with a reasonable explanation.

Lyrx [107]

$40/12 = unit rate simplify = 10/3
$10/3 per book
18 * 10/3 = $60 for 18 copies

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2 years ago

What is the surface area of a tank that has a length of 30 feet, a width of 15 feet?

Lady bird [3.3K]

Answer:

450

Step-by-step explanation:

Multiply the length (30) by the width (15)

30x15=450

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2 years ago

Simplify StartRoot 16 r Superscript 6 Baseline EndRoot

Ksivusya [100]

Answer:

This question is impossible

Step-by-step explanation:

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3 years ago

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Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

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2 years ago

*PLEASE HELP ANSWER* <br> * I WILL GIVE BRAINLIEST *

AlladinOne [14]

Answer:

Union

Step-by-step explanation:

None of the terms in each inequality intersect.

so its not intersection

Its two inequalities put together so its union

4 0

3 years ago

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