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KatRina [158]
3 years ago
12

15 POINTS PLEASE HELP IVE ONLY GOT 10 MINS

Mathematics
2 answers:
algol133 years ago
8 0
I’m pretty sure it’s C
frosja888 [35]3 years ago
7 0
C is what I think it is
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Show whether or not the points with coordinates (1,3) (4,2) (-2,6) are vertices of a right triangle
katen-ka-za [31]

Answer:

No

Step-by-step explanation:

See attached

8 0
3 years ago
Based on the image and stated scale, what is the value of y?
GenaCL600 [577]

Answer:

2.2 inches

Step-by-step explanation:

This is because of the stated scale rate. 1/5 is equal to 20% with that 20% of 11 inches would yield 2.2 inches hope this helps :)

5 0
3 years ago
DETERMINE THE MISSING SIDE
MissTica

Answer:

21 in

Step-by-step explanation:

by the Pythagorean theorem we have

841=400+x^2\\\\x^2=441\\\\x=21

notice that the 841 and the 400 don't go squared because they already squared!

8 0
2 years ago
URGENT!!
slava [35]

Answer:

a. 3/5 liters per hour

b .06 hours

Step-by-step explanation:

Change 40 minutes to a fraction of an hour

40 minutes/ 60 minutes = 2/3 hour

It takes 1 2/3 hours to fill the container

Changing to an improper fraction

(3*1+2)/3 = 5/3 hours

We want liters per hour

1 liter per 5/3 hours

1 ÷ 5/3

Copy dot flip

1 * 3/5 = 3/5 liters per hour

Change 100 ml to liters

We know 1000 ml  = 1 lt

100 ml * 1 lt/1000 ml = .1 lt

So 100 ml = .1 lts

.1 lt * 3/5 lt per hour

Changing the fraction to a decimal

.1 lt * .6 lt per hour

.06 hours

4 0
3 years ago
Evaluate the integral Find the volume of material remaining in a hemisphere of radius 2 after a cylindrical hole ofradius 1 is d
algol13

You want to find the volume inside the hemisphere x^2+y^2+z^2=4 (i.e. inside the sphere but above the plane z=0) and outside the cylinder x^2+y^2=1. Call this region R.

In cylindrical coordinates, we have

\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_1^2\int_0^{\sqrt{4-r^2}}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta

\displaystyle=2\pi\int_1^2 r\sqrt{4-r^2}\,\mathrm dr

\displaystyle=-\pi\int_3^0\sqrt u\,\mathrm du

(where u=4-r^2)

\displaystyle=\pi\int_0^3\sqrt u\,\mathrm du

=\dfrac{2\pi}3u^{3/2}\bigg|_0^3=2\sqrt3\,\pi

7 0
3 years ago
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