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3 years ago

Which of the following functions is graphed below?

Mathematics

1 answer:

geniusboy [140]3 years ago

5 0

Answer:

the answer is c

Step-by-step explanation:

I took it already i think it's that one .

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there were a total of 68 basketball games during the lockout-shortened 2012 NBA season. this season was played over the course o

alisha [4.7K]

The answer is :13...3

3 0

4 years ago

The sum of 1/6, 2/3, and 1/4 is

zimovet [89]

1/6+2/3+1/4
First find the highest common multiple of the denominators=12
Then multiply each term by the quotient of 12/denominator
So...it becomes 2/12+8/12+3/12
Add all the numerators and keep the denominator constant
It becomes (2+8+3)/12=13/12
=1 1/12

5 0

3 years ago

Read 2 more answers

Glasses cost $0.50. You buy 2. How much do you pay?

schepotkina [342]

Answer:

1 dollar

$1.00

Step-by-step explanation:

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2 years ago

In a solar system, two comets pass near the sun, which is located at the origin. Comet E is modeled by quantity y plus 16 end qu

Leya [2.2K]

Explanation:

The equation of comet E is given below as

$\frac{(y+16)^2}{400}+\frac{x^2}{144}=1$

The equation of comet H is modelled below as

$\frac{(y+13)^2}{144}-\frac{x^2}{25}=1$

Given, that the Sun is located at the origin.

This is an equation of an ellipse. So, the path traveled by Comet e is an elliptic path.

Comparing with the Standard equation of ellipse: below,we will have

$\frac{(y-k)^2}{b^2}+\frac{(x-h)^2}{a^2}$

Whise center is

$\begin{gathered} center=(h,k) \\ vertices=(h\pm a,0) \end{gathered}$

By comparing coefficient, we will have

$\begin{gathered} a^2=144 \\ a=12 \\ b^2=400 \\ b=20 \\ k=-16 \\ h=0 \\ (h,k)=(0,-16) \end{gathered}$

Hence,

The vertex of comet E will be

$\begin{gathered} (h\pm a,0) \\ (0+12,0),(0-12,0) \\ (12,0),(-12,0) \end{gathered}$

The vertex of comet E is

$(12,0),(-12,0)$

Part C:

It is the foci forboth comets

$\begin{gathered} cometE: \\ c=\sqrt{b^2-a^2} \\ c=\sqrt{400-144} \\ c=\sqrt{256} \\ c=16 \\ \\ For\text{ comet F:} \\ c=\sqrt{a^2-b^2} \\ c=\sqrt{144+25} \\ c=\sqrt{169} \\ c=13 \end{gathered}$

Hence,

The foci will be

$\begin{gathered} (16\pm16,0) \\ cometE \\ (0,0),(32,0) \\ cometF: \\ (13\pm13,0) \\ (0,0),(26,0) \end{gathered}$

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1 year ago

Which angles are adjacent angles?

Marat540 [252]

Answer:

CBX and FCB

Step-by-step explanation:

trust me

5 0

3 years ago