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geniusboy [140]

2 years ago

7

HELP ME PLEASE!!1 I WILL GIVE !20 POINTS! AND BRAINLIST PLEASE I DON'T UNDERSTAND THIS AND ITS DUE SOON EPLAIN YOUR ANSWERS TOO

PLEASE

2 answers:

Softa [21]2 years ago

7 0

Answer: Question 11 is q/s

Step-by-step explanation:

ryzh [129]2 years ago

4 0

Answer:

Step-by-step explanation:

The two triangles are similar because of their angles. Both of the triangles contain a 90 degree angle, <em>a </em>angle, and 90-<em>a</em> angle.

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Water weighs 62.4 pounds per cubic foot. How much does the water in a tank that measures 7 ft × 4 ft × 9 in, weigh?

andriy [413]

Answer:

<u>1,310 lb</u>

Explanation:

<u>1) Calculate the volume of the water in the tank.</u>

Area of the base of the tank: B = 7 ft × 4 ft = 28 ft²

Height of the tank: H = 9 in = 9 in × 1 ft / 12 in = 0.75 ft

Volume of the tank: V = area of the base × height = B × H = 28 ft² × 0.75 ft = 21 ft³.

<u>2) Calculate the weight of 21 ft³ of water.</u>

Since this is not a chemistry question but a math question, I will not use the fomula of density but set a proportion with one unknown:

62.4 lb / 1 ft³ = x / 21 ft³

Solve for x:

x = 21 ft³ × 64 lb / ft³ = 1,310.4 lb.

So, rounding to the next integer, the water in the tank weighs 1,310 pounds, when it is full.

6 0

3 years ago

Read 2 more answers

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

2 years ago

Read 2 more answers

If |AB| / |DE|=7/5 then what is |AD| / |CD|?DE is parallel to AB

Dmitriy789 [7]

Answer: AD/CD = 2/5

Explanation:

Triangle ABC is similar to triangle DEC on the basis of AA(Angle angle)

Recall, if 2 triangles are similar, it means that the ratio of their corresponding sides are equal. Thus,

AB/DE = BC/EC = AC/DC

From the information given,

AB/DE = 7/5

Thus,

AC/DC = 7/5

AD + CD = AC

(AD + CD)/DC = 7/5

AD/DC + CD/DC = 7/5

AD/DC + 1 = 7/5

AD/DC = 7/5 - 1

AD/DC = 2/5

AD/CD = 2/5

4 0

7 months ago

Can you help me find my mistake

Hunter-Best [27]

Sorry i dont know im not that far in school that's like high school stuff sorry

8 0

3 years ago

Choose all distances that are equivalent to 1/2 mile.

damaskus [11]

Answer:

A, C, and E.

Step-by-step explanation:

I looked up what half a mile was in meters, yards, and feet, and it told me these were the right answers.

6 0

2 years ago