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4 years ago

Quadrilateral ACEG≅ Quadrilateral MNPR Which statement is true?

Mathematics

1 answer:

saul85 [17]4 years ago

4 0

Line CE is congruent to line NP is the true statement.

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Can you please help me?

zubka84 [21]

So the right answer is 9

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Good luck on your assignment

3 0

3 years ago

Find the mean for this set of data 8,3,12,15,7

nika2105 [10]

Answer:

Step-by-step explanation:

If you add them all together and divide by how many numbers there are, you will get the answer.

3 0

3 years ago

Hellllllllllppppppppppp

Levart [38]

You multiply 24 by 6 to get C. $144

8 0

4 years ago

Read 2 more answers

What is the arc measure of ABC in circle P in degrees

Masja [62]

Given:

m∠APB = 74°

To find:

The measure of ACB

Solution:

The measure of the central angle is congruent to the measure of the intercepted arc.

⇒ m(ar AB) = m∠APB

⇒ m(ar AB) = 74°

The complete angle of the circle is 360°.

⇒ m(ar ACB) + m(ar AB) = 360°

⇒ m(ar ACB) + 74° = 360°

Subtract 74° from both sides.

⇒ m(ar ACB) = 286°

The measure of arc ACB is 286°.

7 0

3 years ago

∫∫(x+y)dxdy ,d là miền giới hạn bởi x²+y²=1

igor_vitrenko [27]

It looks like you want to compute the double integral

$\displaystyle \iint_D (x+y) \,\mathrm dx\,\mathrm dy$

over the region D with the unit circle x ² + y ² = 1 as its boundary.

Convert to polar coordinates, in which D is given by the set

D = {(r, θ) : 0 ≤ r ≤ 1 and 0 ≤ θ ≤ 2π}

and

x = r cos(θ)

y = r sin(θ)

dx dy = r dr dθ

Then the integral is

$\displaystyle \iint_D (x+y)\,\mathrm dx\,\mathrm dy = \iint_D r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \int_0^{2\pi} \int_0^1 r^2(\cos(\theta)+\sin(\theta))\,\mathrm dr\,\mathrm d\theta \\\\ = \underbrace{\left( \int_0^{2\pi}(\cos(\theta)+\sin(\theta))\,\mathrm d\theta \right)}_{\int = 0} \left( \int_0^1 r^2\,\mathrm dr \right) = \boxed{0}$

3 0

3 years ago