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3 years ago

Which graph is correct

Mathematics

1 answer:

GrogVix [38]3 years ago

8 0

Answer:

Step-by-step explanation:

Taking the absolute value of anything makes it positive, so you know that the graph of |f(x)| must only have positive.

And since c is the only graph with no negative values c it must be the answer

Also if you compare it with f(x), c it flips the negative parts of f(x) to positive.

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Eli, a pretty smart ape in the Salt Lake City zoo, has correctly picked the winner of the last 7 Super Bowls. Read about Eli's p

Anna [14]

Answer:

There is a 0.78% probability that Eli chooses 7 winners in a row.

Step-by-step explanation:

There were 7 Super Bowls.

In each Super Bowl, there were two teams.

So, the probability that Eli chooses 7 winners in a row is

$P = \frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2}*\frac{1}{2} = \frac{1}{128} = 0.0078$

There is a 0.78% probability that Eli chooses 7 winners in a row.

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3 years ago

Can someone tell me what 800/600+700-2300*32234 equals?

Sever21 [200]

The answer would be -74,137,498.67 when you follow the order of operations. (PEMDAS).

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3 years ago

While you are solving the equation, which choice could be one of your lines? 3 - 2(x - 4) = 5x + 3

Alex73 [517]

It's clear to understand

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3 years ago

Find the zero of each function and state the multiplicity of each zero. Please show all steps.

vodka [1.7K]

Answer:

1. y=(x+3)^3. Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1). Zeros: x=2 multiplicity 2; x=1 multiplicity 1.

3. y=(2x+3)(x-1)^2. Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

Step-by-step explanation:

1. y=(x+3)^3

$y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3$

Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1)

$y=0\\ (x-2)^2(x-1)=0\\ \left \{ {{(x-2)^2=0} \atop {x-1=0}} \right\\ \left \{ {{\sqrt{(x-2)^2} =\sqrt{0} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x-2=0} \atop {x=1}} \right\\ \left \{ {{x-2+2=0+2} \atop {x=1}} \right\\ \left \{ {{x=2} \atop {x=1}} \right.$

Zeros: x=2 multiplicity 2; x=1 multiplicity 1

3. y=(2x+3)(x-1)^2

$y=0\\ (2x+3)(x-1)^2=0\\ \left \{ {{2x+3=0} \atop {(x-1)^2=0}} \right\\ \left \{ {{2x+3-3=0-3} \atop {\sqrt{(x-1)^2} =\sqrt{0} }} \right\\ \left \{ {{2x=-3} \atop {x-1=0}} \right\\ \left \{ {{\frac{2x}{2} =\frac{-3}{2} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x=-\frac{3}{2} } \atop {x=1}} \right.$

Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

4 0

3 years ago

I forgot how to find the area of the inside of the triangle to find the volume.

enot [183]

Maybe use a ruler To find the area and add them together to fin the volume.

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4 years ago