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IgorC [24]
3 years ago
12

Ratio help please, picture inserted.

Mathematics
2 answers:
marta [7]3 years ago
4 0
 The answer is A. b:q
Stells [14]3 years ago
3 0
Im pretty sure it is b
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25x^2+9y^2+72y-81=0. Find the standard form of the equation by completing the square. Then identify and graph each conic.
olasank [31]
          25x² + 9y² + 72y - 81 = 0
                   25x² + 9y² + 72y - 81 + 81 = 0 + 81
                                   25x² + 9y² + 72y = 81
                           25x² + 9(y² + 8y + 16) = 81 + 9(16)
                   25x² + 9(y² + 4y + 4y + 16) = 81 + 144
           25x² + 9(y(y) + 4(y) + 4(y) + 4(4)) = 225
                  25x² + 9(y(y + 4) + 4(y + 4)) = 225
                            25x² + 9(y + 4)(y + 4) = 225
                                     <u>25x²</u> + <u>9(y + 4)²</u> = <u>225</u>
                                     225         225       225
                                         <u> x²</u> + <u>9(y + 4)²</u> = 1
                                         25        25

The conic section is an ellipse.
8 0
3 years ago
Please help! Will mark brainliest
iris [78.8K]
5 x 10^-2, 50%, 1/3, 0.6
Second option
5 0
3 years ago
The graph below shows a quadratic function f(x) and an absolute value function g(x)
Sophie [7]

Answer:

The x-values that are approximate solution of the equation f(x) = g(x) are x_{1} \approx -3.5 and x_{2}\approx 5, respectively.

Step-by-step explanation:

From graph we infer that the x-values that represents a solution to the system of equations corresponds to those values of x in which both functions coincide. As we see, there are two x-values for f(x) = g(x), which are described below:

x_{1} \approx -3.5 and x_{2}\approx 5

The x-values that are approximate solution of the equation f(x) = g(x) are x_{1} \approx -3.5 and x_{2}\approx 5, respectively.

6 0
3 years ago
Find (f • g) when f(x) = x^2 + 5x + 6 and g(x) = 1/x+3
Nikitich [7]

Answer:

option D

Step-by-step explanation:

f(x) = x^2 + 5x + 6

g(x)= \frac{1}{x+3}

(fog)(x) = f(g(x))

Plug in g(x) in f(x)

We plug in 1/x+3 in the place of x  in f(x)

f(g(x))= f(\frac{1}{x+3})= (\frac{1}{x+3})^2 + 5(\frac{1}{x+3}) + 6

To simplify it we take LCD

LCD is (x+3)(x+3)

\frac{1}{(x+3)(x+3)}+5\frac{1*(x+3)}{(x+3)(x+3)}+\frac{6(x+3)(x+3)}{(x+3)(x+3)}

\frac{1}{x^2+6x+9}+\frac{(5x+15)}{x^2+6x+9}+\frac{6x^2+36x+54}{x^2+6x+9}

All the denominators are same so we combine the numerators

\frac{1+5x+15+6x^2+36x+54}{x^2+6x+9}

\frac{6x^2+41x+70}{x^2+6x+9}

Option D is correct

8 0
3 years ago
Read 2 more answers
What is the answer please help no links I will report
fomenos
I found the answer. hehehehehehehehrheehrhrh

5 0
3 years ago
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