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3 years ago

What is an equation of a circle with center (2,7) and radius 4?

Mathematics

1 answer:

goblinko [34]3 years ago

8 0

Equation of a circle:

(x - h)^2 + (y - k)^2 = r^2

In this case:

<span>center (2,7) and radius 4 so h = 2, k = 7 and r = 4

</span>Equation:

(x - 2)^2 + (y - 7)^2 = 4^2

(x - 2)^2 + (y - 7)^2 = 16

Hope it helps.

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Directions: Solve for x.<br> Please help

aksik [14]

Answer:

Step-by-step explanation:

Since the whole right side of that triangle is equal to 28 and the lower portion is 21, then we know the upper portion is 7. Set up a proportion using those sides and the fact that the 2 triangles are similar:

$\frac{7}{21} =\frac{x}{18}$ and cross multiply to get

21x = 126 and

x = 6

7 0

2 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

7 0

2 years ago

PLEASE HELP ME ASAP I NEED TO DO THIS

yarga [219]

Answer:

1. D

Step-by-step explanation:

8 0

2 years ago

Suppose x and y vary together such that x is 7 times as large as y. write a formula that defines x in terms of y.

Ainat [17]

Since y is less than x by 7 times $\frac{x}{7}=y$ so in terms of x we write
x=7y

4 0

3 years ago

What are the inputs of the following function?

igor_vitrenko [27]

Answer:

A. {-4, -3, 7, 8}

Step-by-step explanation:

The ordered pairs representing a function are always written ...

(input, output)

<h3>Inputs</h3>

The set of inputs for the given function is the list of first-numbers of the ordered pairs. Those numbers are -3, -4, 8, 7. When we express them as a set, we like to have the elements of the set in increasing order:

inputs = {-4, -3, 7, 8} . . . . . . matches the first choice

7 0

1 year ago