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3 years ago

What are the center and radius of a circle defined by the equation x^2+y^2- 6x+ 8y+ 21=0?

Mathematics

2 answers:

KengaRu [80]3 years ago

8 0

Answer:

It is B, Thank me later

Step-by-step explanation:

Natali5045456 [20]3 years ago

4 0

You have to complete the square. When you do this you get an equation that looks like this: (x-3)^2 + (y+4)^2 = 4. So your center is (3, -4) and your radius is 2. That looks like B to me!!!

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Can someone please help me with writing a flowchart proof?

Keith_Richards [23]

Answer:

∠AZ = ∠BX

The two triangles are equivalent triangles

Step-by-step explanation:

Flowchart proofs are organized with boxes and arrows. The statements given are written in the box and the reason of that statement is written just below the box. Every statement if flowchart proof complements other statement.

∠AZ = ∠BX

The two triangles are equivalent triangles.

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2 years ago

Select the correct answer.

liubo4ka [24]

Answer:

Step-by-step explanation:

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3 years ago

The coordinates of the vertices of ΔABC are A(−5, 1), B(−5, −1), and C(−1, −1). Find the side lengths and the angle measures.

givi [52]

Answer:

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Step-by-step explanation:

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2 years ago

How do I solve this?

Solnce55 [7]

Answer:

Step-by-step explanation:

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2 years ago

Read 2 more answers

The marketing director of a large department store wants to estimate the average number of customers who enter the store every f

Dahasolnce [82]

Answer:

$36.5674\leq x'\leq61.4326$

Step-by-step explanation:

If we assume that the number of arrivals is normally distributed and we don't know the population standard deviation, we can calculated a 95% confidence interval to estimate the mean value as:

$x-t_{\alpha /2}\frac{s}{\sqrt{n} } \leq x'\leq x-t_{\alpha /2}\frac{s}{\sqrt{n} }$

where x' is the population mean value, x is the sample mean value, s is the sample standard deviation, n is the size of the sample, $\alpha$ is equal to 0.05 (it is calculated as: 1 - 0.95) and $t_{\alpha /2}$ is the t value with n-1 degrees of freedom that let a probability of $\alpha/2$ on the right tail.

So, replacing the mean of the sample by 49, the standard deviation of the sample by 17.38, n by 10 and $t_{\alpha /2}$ by 2.2621 we get:

$49-2.2621\frac{17.38}{\sqrt{10} } \leq x'\leq 49+2.2621\frac{17.38}{\sqrt{10} }\\49-12.4326\leq x'\leq 49+12.4326\\36.5674\leq x'\leq61.4326$

Finally, the interval values that she get is:

$36.5674\leq x'\leq61.4326$

8 0

3 years ago