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2 years ago

Find an equation of the circle that has center (-4, 1) and passes through (3, -5)

Mathematics

1 answer:

Lemur [1.5K]2 years ago

5 0

The general equation for a circle with center (a,b) and squared radius k is

$(x-a)^2 + (y-b)^2 = k$

Here we have

$(x- - 4)^2 + (y -1)^2 = k$

and the constant is fixed by substituting in the point we know (x,y)=(3,-5)

$(x + 4)^2 + (y-1)^2 = (3+4)^2 + (-5-1)^2 = 85$

Answer:

$(x + 4)^2 + (y-1)^2 = 85$

You might be interested in

At what value of x does the graph of the following function f(x) have a vertical asymptote? f(x)=1/x+3

ratelena [41]

Answer:

Step-by-step explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined.

Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non zero for this value then it is a vertical asymptote.

x + 3 = 0 ⇒ x = - 3 ← equation of vertical asymptote

5 0

3 years ago

Read 2 more answers

PLEASE HELP! NEED TO SUBMIT ASAP!!!!!!!! PLEASE, THANK YOU IN ADVANCE!!!!! 11PTS

Leni [432]

This is a table with 6 lines.
In each line, the number in the first column is the 'x' value.

All you have to do on each line is ...

--- substitute the 'x' value in (100 + 23x), simplify it,
and write the result in the middle column

then

--- substitute the 'x' number in 90(1.2ˣ) , simplify it,
and write the result in the last column.

On the first line, x=0.

100 + 23x = 100 + 0 = 100. Write 100 in the middle column.

90(1.2ˣ) = 90(1) = 90. Write 90 in the last column.

Then go on to the second line, where x=1.

You'll make it.

7 0

3 years ago

Please help solve for x..

aalyn [17]

Answer:

x = 1 or x = -1

Step-by-step explanation:

Given equation:

$x^{100}-4^x \cdot x^{98}-x^2+4^x=0$

Factor out -1:

$\implies -1(-x^{100}+4^x \cdot x^{98}+x^2-4^x)=0$

Divide both sides by -1:

$\implies -x^{100}+4^x \cdot x^{98}+x^2-4^x=0$

Rearrange the terms:

$\implies 4^x \cdot x^{98}-4^x-x^{100}+x^2=0$

$\textsf{Apply exponent rule} \quad a^{b+c}=a^b \cdot a^c \quad \textsf{to }x^{100}:$

$\implies 4^x \cdot x^{98}-4^x-x^{98}x^2+x^2=0$

Factor the first two terms and the last two terms separately:

$\implies 4^x(x^{98}-1)-x^2(x^{98}-1)=0$

Factor out the common term $(x^{98}-1)$ :

$\implies (4^x-x^2)(x^{98}-1)=0$

<u>Zero Product Property</u>: If a ⋅ b = 0 then either a = 0 or b = 0 (or both).

Using the <u>Zero Product Property</u>, set each factor equal to zero and solve for x (if possible):

$\begin{aligned}x^{98}-1 & = 0 & \quad \textsf{or} \quad \quad4^x-x^2 & = 0 \\x^{98} & =1 & 4^x & = x^2 \\x & = 1, -1 & \textsf{no}& \textsf{ solutions for } x \in \mathbb{R}\end{aligned}$