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frozen [14]
3 years ago
15

PLEASE HELP THIS IS URGENT!

Mathematics
1 answer:
andriy [413]3 years ago
3 0

Answer:

D

Step-by-step explanation:

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In the triangle XYZ, IF WZ=24, then WY is:
lukranit [14]
WZ is congruent to WY based on the picture, so WY is also 24.
6 0
3 years ago
A park district has 25 elm trees and 20 oak trees to be planted. They are separating the trees for different areas. They want ea
Vlada [557]

This question belongs to greatest common factor.

Given, there are two types of trees.

25 elm trees and 20 oak trees.

Let's find the factors of 25 and 20

25 = 5 * 5

20 = 5* 4

The common factor between both the numbers is 5.

Hence, there should be 5 areas so that the trees can be planted equally.

Answer = 5 areas.

6 0
3 years ago
Read 2 more answers
Please answer the question and explain your answer. I will be marking brainiest with the correct answer and best explanation.
qaws [65]

Answer:

28.26

Step-by-step explanation:

A = r^2

So the equation is. Area of circle = pi(3.14) * radius(3) to the power of 2 (^2)

A = 3.14(3)^2

A = 3.14(9)

A = 28.26

6 0
3 years ago
A) Table Mountain is 1085 meters above sea level is it a positive or negative integer ​
leonid [27]

Answer:

It is positive

Step-by-step explanation:

because of geomatrical graph

6 0
3 years ago
Find the area of the surface. the part of the hyperbolic paraboloid z = y2 − x2 that lies between the cylinders x2 + y2 = 4 and
Nadusha1986 [10]
Call the surface S; then the area of S is given by the surface integral

\displaystyle\iint_S\mathrm d\mathbf S

Parameterize the surface by

\mathbf r(u,v)=\begin{cases}x(u,v)=u\cos v\\y(u,v)=u\sin v\\z(u,v)=u^2\cos^2v-u^2\sin^2v=u^2\cos2v\end{cases}

with 2\le u\le4 and 0\le v\le2\pi. The surface element is given by

\mathrm d\mathbf S=\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv
\mathrm d\mathbf S=u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv

So the area is

\displaystyle\iint_S\mathrm d\mathbf S=\int_{v=0}^{v=2\pi}\int_{u=2}^{u=4}u\sqrt{1+4u^2}\,\mathrm du\,\mathrm dv
=\displaystyle2\pi\int_{w=17}^{w=65}\sqrt w\,\frac{\mathrm dw}8

where w=1+4u^2\implies\mathrm dw=8u\,\mathrm du

=\displaystyle\frac\pi4\frac23w^{3/2}\bigg|_{w=17}^{w=65}
=\dfrac\pi6\left(65^{3/2}-17^{3/2}\right)\approx237.69
8 0
3 years ago
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