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2 years ago

It would be appreciated if there could be an explanation along the answer

Mathematics

1 answer:

soldi70 [24.7K]2 years ago

8 0

Answer:

i think probably 80°

Step-by-step explanation:

PQ = RS

2x+3 = 4x-3

x = 3

PTQ = RTS

9y-3x-1 = 5x+7y-5

2y-8x = -4

y-4x = -2

y-4(3) = -2

y-12 = -2

y = -2+12

y = 10

so PTQ = 9y-3x-1

9(10)-3(3)-1

90-9-1

80°

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The height of a pyramid is doubled, but its length and width are cut in half. What is true about the volume of the new pyramid?

Aleksandr [31]

Answer:

The new pyramid has a volume that is half the volume of the original pyramid.

Step-by-step explanation:

The volume of a pyramid is given by the following equation:

$V = \frac{h*l*w}{3}$

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Modified volume:

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$h = 2h, l = 0.5l, w = 0.5w$

$V_{M} = \frac{2h*0.5l*0.5l}{3} = \frac{0.5*h*l*w}{3} = 0.5\frac{h*l*w}{3}$ = 0.5V[/tex]

That is

The new pyramid has a volume that is half the volume of the original pyramid.

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3 years ago

Quadrilateral A'B'C'D is the image of quadrilateral ABCD under a translation.

Temka [501]

Answer: 1 unit to the left, and 3 units down

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

I need help please and thank you

Brums [2.3K]

Well I THINK triangle ABC is just a stretched out version of triangle DEF.

To find the area of a triangle, you multiply base and height, and divide by 2.

Area of DEF is 6, so that means that (4 * x)/2 = 6.

6*2 = 12....

4 * x = 12........

4 * 3 = 12.

x = 3.

Comparing DEF to ABC: ABC's base is 3 times as long as DEF's. So that means the height must be three times as long.

3*3 = 9.

(12 * 9)/2 =54

5 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%7B%20%20%5Cpi%20%7D%20%5Ccos%28%20%5Ccot%28x%29%20%20%20%20-%20%2

Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

but the integrand is even, so this is really just

$\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

Substitute x = 1/2 arccot(u/2), which transforms the integral to

$\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du$

There are lots of ways to compute this. What I did was to consider the complex contour integral

$\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz$

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

$\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}$

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

$\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}$

and it follows that

$\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}$

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2 years ago

A doctor told his patient to drink 3 full cups and 1/2 of a cup of medicine over a week. If each full cup was 4 1/2 pints,how mu

fenix001 [56]

-3x4.5=13.5. 4.5x0.5=2.25

-13.50
+2.25
_____
15.75 pints

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3 years ago