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2 years ago

Write the standard form of the equation of the circle with radius 7 and center (5,−7).

Mathematics

1 answer:

olya-2409 [2.1K]2 years ago

4 0

Answer:

$(x-5)^2+(y+7)^2=49$

Step-by-step explanation:

Use $(x-h)^2+(y-k)^2=r^2$ for the equation of a circle in standard form with radius $r$ and center $(h,k)$ . Therefore, the equation of the circle in standard form is $(x-5)^2+(y+7)^2=49$

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Which of these is a complex number?

Ostrovityanka [42]

Answer:

Step-by-step explanation:

That would be B.

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The √(-7/3) is usually written as √(7/3)i where i is √-1, so the complex number is 8/3 + √(7/3)i.

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3 years ago

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What is the of the function?

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I think the answer is (-3,1) but I'm not 100% sure.

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2 years ago

I need helping answering 15. I need to create a table and graph the function.

Alekssandra [29.7K]

ANSWER and EXPLANATION

We want to order the functions from widest to narrowest:

$\begin{gathered} y=3x^2 \\ y=2x^2 \\ y=4x^2 \end{gathered}$

To do this, we have to plot the graphs of the functions by using a table of values.

Let us find the values of the functions for values of x = -2, 0, 2

For the first function:

$\begin{gathered} y=3x^2 \\ \Rightarrow x=-2: \\ y=3(-2)^2=3*4 \\ y=12 \\ \Rightarrow x=0: \\ y=3(0)^2=3*0 \\ y=0 \\ \Rightarrow x=2: \\ y=3(2)^2=3*4 \\ y=12 \end{gathered}$

Hence, its table is:

For the second function:

$\begin{gathered} y=2x^2 \\ \Rightarrow x=-2: \\ y=2(-2)^2=2*4 \\ y=8 \\ \Rightarrow x=0: \\ y=2(0)^2=2*0 \\ y=0 \\ \Rightarrow x=2: \\ y=2(2)^2=2*4 \\ y=8 \end{gathered}$

Hence, its table is:

For the third function:

$\begin{gathered} y=4x^2 \\ \Rightarrow x=-2: \\ y=4\left(-2\right)^2=4*4 \\ y=16 \\ \Rightarrow x=0: \\ y=4(0)^2=4*0 \\ y=0 \\ \Rightarrow x=2: \\ y=4(2)^2=4*4 \\ y=16 \end{gathered}$

Hence, its table is:

Now, let us plot the graphs of the functions:

Therefore, from the graph, we see that the order of the functions from widest to narrowest is:

$y=2x^2;y=3x^2;y=4x^2$

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10 months ago

6x-5y=15. X=y+3. X= ?y=?

sukhopar [10]

Answer:

x=0

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x=y+3

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3 years ago

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denis-greek [22]

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3 years ago

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