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4 years ago

Create a function with roots x = 3 and x = 2 and x = -3

Mathematics

1 answer:

WITCHER [35]4 years ago

6 0

That would be
f(x) = (x - 2)(x - 3)(x + 3)

expanded this is f(x) = (x - 2)(x^2 - 9)
= x^3 - 2x^2 - 9x + 18

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Alexis wants to make a paperweight at pottery class. He designs a pyramid-like mode l with a base area of 100 square centimeters

olga55 [171]

Answer:

$\frac{3}{2}$ g/cm²

Step-by-step explanation:

First, we need to find the volume of the prism. The volume of a square pyramid is V = Bh/3

B = 100 cm²

h = 6 cm

V = 100 cm² * 6 cm /3

V = 600 cm³ /3

V = 200 cm³

Next, we need to solve for the density based on the desired volume and mass.

m = mass = 300 g

V = volume

density = m/V

density = 300 g / 200 cm³

density = 3/2 g/cm³

The lowest possible density of the material to make the paperweight would be $\frac{3}{2}$ g/cm²

4 0

3 years ago

The ratio of boys to girls is 4:3 .T here are 35 students in a class how many students are boys ?? Pleaseee help

evablogger [386]

There are 20 boys in the class. The ratio is 4:3 so that means 4 out of 7 are boys and 3 out of 7 are girls. Divide 35 by 7 and you get 5. 5 times 4 is 20. 5 times 3 is 15. add them together and you have 35 students in total. So 20 students are boys and 15 students are girls.

8 0

3 years ago

Last night, the temperature fell from 0°F to Negative 13 and one-fifth°F in4 and two-fifths hours. Alan calculated the average t

USPshnik [31]

Negative 3.14
Times that by Pi which is 3.14 and then multiply by the base number and you will get the correct answer

5 0

3 years ago

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Evaluate Dx / ^ 9-8x - x2^

Solnce55 [7]

It depends on what you mean by the delimiting carats "^"...

Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for $\sqrt x$ .

In that case, you want to find the antiderivative,

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}$

Complete the square in the denominator:

$9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2$

Now substitute $x+4=5\sin y$ , so that $\mathrm dx=5\cos y\,\mathrm dy$ . Then

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy$

which simplifies to

$\displaystyle\int\frac{5\cos 
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy$

Now, recall that $\sqrt{x^2}=|x|$ . But we want the substitution we made to be reversible, so that

$x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)$

which implies that $-\dfrac\pi2\le y\le\dfrac\pi2$ . (This is the range of the inverse sine function.)

Under these conditions, we have $\cos y\ge0$ , which lets us reduce $\sqrt{\cos^2y}=|\cos y|=\cos y$ . Finally,

$\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C$

and back-substituting to get this in terms of $x$ yields

$\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C$

4 0

3 years ago

-10÷0.22 what is the answer

Juliette [100K]

Answer:

-45.4545454545 *45 repeating*

Step-by-step explanation:

8 0

2 years ago

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