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SIZIF [17.4K]
3 years ago
15

Write the ratio in two different ways. (6.PA.1)

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
8 0
Gkiydwsgnlliuhfwwxbjiurd
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Part 1: Write the equation of the line that passes through the points (–1, –4) and (2, 5).
jeyben [28]
Y - (-4) = (5 - (-4))/(2 - (-1)) (x - (-1))
y + 4 = (5 + 4)/(2 + 1) (x + 1)
y + 4 = 9/3 (x + 1)
y + 4 = 3(x + 1)
y + 4 = 3x + 3
y = 3x + 3 - 4
y = 3x - 1

5 0
4 years ago
Read 2 more answers
Z=5x-11y ?<br><br> and-2-5v &lt; 8
velikii [3]

Answer:

Step-by-step explanation: v> -2 the > has a line under it

3 0
4 years ago
he Pearson family is on vacation in Alaska. Since Mr. Peterson has always wanted to see Mt. McKinley, they are taking a scenic t
Aloiza [94]

Answer:

Mr. Peterson

Step-by-step explanation:

u and one

6 0
3 years ago
Solve 2x^3+7x^2-2x-1=o​
Mashcka [7]

Answer:

x=\sqrt{3}-2\\x=-\sqrt{3}-2\\x=\frac{1}{2}

Step-by-step explanation:

This is a hard one

We have to use the rational root theorem

2x^3+7x^2-2x-1 = 0

We have to find all the factors of a and d and put them in a fraction

a=2\\d=-1\\\frac{1}{1,2}

We then plug them into the equation to see if any of them work

The equation isn't true when plugging 1, but is true when plugging in 1/2

factored form of 1/2 is (2x-1)

Then we divide the original equation by (2x-1) (you can use synthetic division or long division, it would be hard to type out the process for that) to get x^2+4x+1

So now the equation is (2x-1)(x^2+4x+1)

Solve the second half of this equation using the quadratic formula to get

\sqrt{3}-2 and -\sqrt{3}-2

We already know the solution for the first half of the equation (1/2)

So the final answers are:

x=\sqrt{3}-2, -\sqrt{3}-2, \frac{1}{2}

8 0
3 years ago
The radius of a sphere increases at a constant rate of 2 com/min at the time when the volume of the sphere is 40 cm^3ç What is t
ExtremeBDS [4]

Answer:

\frac{dV}{dt}=525 cm^{3}/min

Step-by-step explanation:

The volume of a sphere is:

V=\frac{4}{3}\pi r^{3} (1)

We know that:

  • dr/dt = 2 cm/min (increasing rate of radius)
  • V = 40 cm³

If we take the derivative of (1) with respect of time t, we ca n find the rate of increase of the volume.

\frac{dV}{dt}=\frac{4}{3}\pi 3r^{2}\frac{dr}{dt}=4\pi r^{2}\frac{dr}{dt} (2)

We also know that the volume is 40 cm³, then using the (1) we can get the radius at this value.

Solving (1) for r, we have:

r=\left(\frac{3\cdot V}{4\pi}\right)^{1/3}=\left(\frac{3\cdot 400}{4\pi}\right)^{1/3}=4.57 cm

Finally dV/dt will be:

\frac{dV}{dt}=4\pi (4.57)^{2}\cdot 2=525 cm^{3}/min

I hope it helps you!

6 0
3 years ago
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