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3 years ago

15

Write the equation in standard form, then identify center and radius of circle.

1 answer:

Natalka [10]3 years ago

6 0

Answer:

Step-by-step explanation:

Given

x^2 + y^2 - 16y - 36 = 0

x^2 + y^2 - 16y = 36

x^2 + y^2 - 16y + (16/2)^2 = 36 + 8^2

x^2 + (y - 8)^2 = 64 + 36

x^2 + (y - 8)^2 = 100

The center is at (0,8) and the radius is 10

Try and follow it from the graph below.

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HELP!!!<br> HALP MEEEEEEEE

goldfiish [28.3K]

Answer:

Bryce's unit rate in miles around the track is 3 rate per mile . Now, as we know the formula for speed i.e. So, Bryce's unit rate in miles around the track is 3 rate per mile .

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2 years ago

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Find out the answers to this

meriva

Answer:

$\tan \theta = - \frac{1}{5} = - 0.2$

$\cos \theta = 0.98$

$\sin \theta = - 0.196$

Step-by-step explanation:

It is given that $\cot \theta = - 5$ and $\theta$ is in the fourth quadrant.

So, only $\cos \theta$ will have positive value and $\sin \theta$ , $\tan \theta$ will have negative value.

Now, $\cot \theta = - 5$

⇒ $\tan \theta = \frac{1}{\cot \theta} = -\frac{1}{5}$ (Answer)

We know, that $\sec^{2} \theta - \tan^{2} \theta = 1$

⇒ $\sec \theta = \sqrt{1 + \tan^{2} \theta } = \sqrt{1 + (- \frac{1}{5} )^{2} } = 1.019$

{Since, $\cos \theta$ is positive then $\sec \theta$ will also be positive}

⇒ $\cos \theta = \frac{1}{\sec \theta} = \frac{1}{1.0198} = 0.98$ (Answer)

We know, that $\csc^{2} \theta - \cot^{2} \theta = 1$

⇒ $\csc \theta = - \sqrt{1 + \cot^{2} \theta } = - \sqrt{1 + (- 5 )^{2} } = - 5.099$

{Since, $\sin \theta$ is negative then $\csc \theta$ will also be negative}

⇒ $\sin \theta = \frac{1}{\csc \theta} = \frac{1}{- 5.099} = - 0.196$ (Answer)

4 0

3 years ago

Find the interquartile range of 17, 23, 8, 5, 9, 16, 22, 11, 13, 15, 17, 18

Yuri [45]

8.25

Hope I helped! ( Smiles )

4 0

3 years ago

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Determine the number of possible triangles, ABC, that can be formed given angle A = 30°, a = 4, and b = 6.

Sophie [7]

Answer:

Step-by-step explanation:

Alright, lets get started.

using Sine Law,

$\frac{sinA}{a}=\frac{sinB}{b}$

$\frac{sin30}{4}=\frac{sinB}{6}$

$sinB=0.75$

$angle B = 48.6$

Another angle will be

$angle B' = 180-48.6 = 131.4$

considering angle B, angle C = $180 - (48.6+30)=101.4$

considering angle B', angle C' = $180-(131.4+30)=18.6$

$\frac{sinA}{a}=\frac{sinC}{c}$

$\frac{sin30}{4}=\frac{sin101.4}{c}$

$c = 7.84$

Similarly, finding c'

$\frac{sinA}{a}=\frac{sinC'}{c'}$

$\frac{sin30}{4}=\frac{sin18.6}{c'}$

$c'=2.55$

Hence two triangles are possible with below details: : Answer

A = 30, B = 48.6, C = 101.4, c = 7.84

A = 30, B' = 131.4, C' = 18.6, c' = 2.55

Hope it will help :)

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2 years ago

What is the equivalent mixed number to 23/4

lora16 [44]

5 3/4
5 remainder 3
Is the mixed number

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2 years ago

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