4 years ago

Ill give 5 star rate and brain answer

1 answer:

5 0

The solutions are where the lines intersect

solution to p(x) and f(x) is (6,2)...because this is where they intersect

solutions to f(x) are : (6,2) and (3,6)

solution to p(x) and g(x) is (3,6)

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Find a so that the point (-1, 2) is on the graph of f(x)=ax^2+5.

Gnesinka [82]

We are asked to determine the value of a such that the function f(x) = ax^2 + 5 is fit for the point given (-1,2). In this case, we substitute 2 to y and -1 to x. The result is then 2 = a*(-1)^2 + 5 ; 2 = a + 5; a is then equal to -3

4 0

4 years ago

(MULTIPLE CHOICE QUESTION)

weeeeeb [17]

Answer:

Transitive property is used in step 4.

Step-by-step explanation:

Transitive property states that if

$a = b \\b = c\\ then\\a = c$

Here we have

$\angle 5 \cong \angle 1\\\angle 1 \cong \angle 3\\then\\\angle 3 \cong \angle 5$

3 0

3 years ago

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Help me...

katen-ka-za [31]

<h2>>> Answer </h2>

________

3y + 1 = 10

3y = 10 - 1

3y = 9

y = 9 ÷ 3

y = 3

6 0

3 years ago

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The equation 9y-12x=36 is written in standard form what is the first step of the equation when writing an equivalent equation wh

Doss [256]

I hope this helps you

7 0

3 years ago

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Point Q'Q ′ Q, prime is the image of Q(0,6)Q(0,6)Q, left parenthesis, 0, comma, 6, right parenthesis under the translation (x,y)

Marina86 [1]

Q is located at (0,6)

The translation rule is $(x,y) \to (x+7,y-5)$ which says to add 7 to the x coordinate and subtract 5 from the y coordinate. Doing that to (0,6) moves it to (7,1) which is where point Q' is located.

In other words, if you shift point Q(0,6) seven units to the right and five units down, then it arrives at Q ' (7,1)

<h3>Answer: (7, 1)</h3>

8 0

4 years ago