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3 years ago

I WILL MARK CROWN JUST NEED HELP ON THIS MAKE SURE RIGHT ANSWER

Mathematics

1 answer:

romanna [79]3 years ago

5 0

hope this will help u.......

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Part F Does this mean that triangle 1 is a right triangle? Why or why not?

erik [133]

Answer:

i'm doing the same question but i think it is a right triangle but in a different position.

Step-by-step explanation:

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3 years ago

Please help meeee!! :(

kirill115 [55]

The function for x=3 is 1

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3 years ago

Today 67 students are going on a

Fudgin [204]

I know this is a little weird but the answer is 1.34 buses.

You need to divide 67 by 50.

Hope I helped and good luck!

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3 years ago

Read 2 more answers

What is the cube root of -729a^9b^6? -9a^3b^2 -9a^2b^ -8a^3b^2 -8a^2b^3

timurjin [86]

Answer:

$= 9a^3b^2$

Step-by-step explanation:

$\sqrt[3]{-729a^9b^6} =$

$= ((-9)^3a^9b^6)^{\frac{1}{3}$

$= (-9)^{3 \times \frac{1}{3}} \times a^{9 \times \frac{1}{3}} \times b^{6 \times \frac{1}{3}}$

$= (-9)^{1} \times a^{3} \times b^{2}$

$= -9a^3b^2$

7 0

2 years ago

1. The mechanics at Lincoln Automotive are reborning a 6 in deep cylinder to fit a new piston. The machine they are using increa

Firdavs [7]

Answer:

$0.0239\frac{in^{3}}{min}$

Step-by-step explanation:

In order to solve this problem, we must start by drawing a diagram of the cylinder. (See attached picture)

This diagram will help us visualize the problem better.

So we start by determining what data we already know:

Height=6in

Diameter=3.8in

Radius = 1.9 in (because the radius is half the length of the diameter)

The problem also states that the radius will increase on thousandth of an inch every 3 minutes. We can find the velocity at which the radius is increasing with this data:

$r'=\frac{1/1000in}{3min}$

which yields:

$r'=\frac{1}{3000}\frac{in}{min}$

with this information we can start solving the problem.

First, the problem wants us to know how fast the volume is increasing, so in order to find that we need to start with the volume formula for a cylinder, which is:

$V=\pi r^{2}h$