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3 years ago

What is the radius of a circle whose equation is x2 + (y – 8)2 = 25?

Mathematics

2 answers:

Dafna11 [192]3 years ago

8 0

Use photo math :) and you will get your answer

Ann [662]3 years ago

5 0

Answer:

r = 5

Step-by-step explanation:

Compare this equation to the standard form

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

Through this comparison we see that h = 0, k = 8 and r = 5.

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How put a number as a percentage

kirill115 [55]

Answer:

You need more information. But typically multiplly by 100 so .1 would be 10 %

Step-by-step explanation:

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2 years ago

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Kindly solve this question 6 too !!

Shkiper50 [21]

Answer:

$x=1$

Step-by-step explanation:

$~~~~~x = \dfrac{1}{2 - \dfrac 1{2-\dfrac{1}{2-x}}}\\\\\\\implies x = \dfrac{1}{2 - \dfrac 1{\dfrac{4-2x-1}{2-x}}}\\\\\\\implies x= \dfrac{1}{2 - \dfrac{2-x}{3-2x}}\\\\\\\implies x = \dfrac{1}{\dfrac{6-4x-2+x}{3-2x}}\\\\\\\implies x = \dfrac{3-2x}{4-3x}\\\\\\\implies 4x -3x^2 = 3-2x\\\\\\\implies 3x^2 -4x +3-2x = 0\\\\\\\implies 3x^2 -6x +3 = 0\\\\\\\implies 3(x^2 -2x +1) =0\\\\\\\implies x^2 -2x +1 = 0\\\\\\\implies (x-1)^2 = 0\\\\\\\implies x -1 = 0\\\\\\\implies x = 1$

7 0

1 year ago

Any REAL help on #16 above??? Can someone check my answer?

babymother [125]

Answer:

The answer to your question is:th first option is correct.

Step-by-step explanation:

Here we have and hyperbola with center (0, 1), and the hyperbola is horizontal because x² is positive.

Equation

y - k = ± $\frac{b}{a} (x - h)$

Process

Find a, b

a² = 9

a = 3

b² = 5

b = √5

h = 0 and k = 1

Substitution

y - 1 = ± $\frac{√5}{3} (x - 0)$

Equation 1

y = $\frac{\sqrt{5} }{3} x + 1$

Equation 2

y = - $\frac{\sqrt{5} }{3} x + 1$

3 0

2 years ago

How do I solve 2f+10?

Scorpion4ik [409]

First of all, you want the variable to be isolated, so in order to do that, you divide in both sides. Ten divided by two is five. So f = 5.

2f + 10
--- ---
2 2

10 divided by 2 = 5

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3 years ago

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Need help with a simple problem 2^x=3^x+1

KiRa [710]

For $x\ \textgreater \ 0$ , $2^x\ \textless \ 3^x$ , all the more $2^x\ \textless \ 3^x+1$ , therefore no solution.

For $x=0$ , we have $2^0=3^0+1 \Rightarrow 1=1+1 \Rightarrow 1=2$ . Obviously that' false.

For $x\ \textless \ 0$ both $2^x$ and $3^x$ , where $2^x\ \textgreater \ 3^x$ , belong to the range $(0,1)$ , then $3^x+1$ belong to the range $(1,2)$ .
Those ranges don't have a common part, therefore, again, no solution.

So, the equation $2^x=3^x+1$ doesn't have a solution.

4 0

2 years ago