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The graphs below have the same shape. What is the equation of the red graph?

Sveta_85 [38]

we have

$g(x)=4-x^{2}$

the vertex of the function g(x) is the point $(0,4)$

the vertex of the function f(x) is the point $(0,0)$

so

the rule of the translation of g(x) to f(x) is

(x,y)------> (x,y-4)

that means

the translation is $4$ units down

therefore

the equation of the function f(x) is

$f(x)=g(x)-4\\f(x)=(4- x^{2})-4 \\f(x)=- x^{2}$

therefore

<u>the answer is</u>

$f(x)=- x^{2}$

8 0

3 years ago

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Question 4 of 15

melisa1 [442]

“C” Os the correct answer

5 0

3 years ago

) Ramjanam makes an arrangement for accommodation of 800 guests in his son's marriage

antiseptic1488 [7]

I don’t know if I have any

8 0

3 years ago

Sammie read at the rate of 25 pages per hour. Alex read at the rate of 1/2 of a pages per minute. Who was reading faster? By how

kompoz [17]

Answer:

Alex and 5 pages

Step-by-step explanation:

Sammie 25 pages per hour

Alex 1/2 pages per minute

You do 1/2x60=30 because you can find out how many pages he can read per hour. Which is 30 pages per hour. So Alex can read 30 pages per hour and Sammie can only read 25 pages per hour. To find how many pages he reads faster per hour, you have to take 30 and subtract it by 25 to get 5 extra pages. So Alex reads 5 pages faster per hour. In the end, Alex reads faster than Sammie by 5 more pages per hour.

Hope this helps! Please give brainliest.

3 0

3 years ago

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Help I don’t understand at all

zloy xaker [14]

Answer:

$1)\ \ 4h^2-13h+6\\2)\ \ 7x^3y^2-x^2y+1\\3)\ \ -7n+2\\4)\ \ -8m+4$

Step-by-step explanation:

1.

Simplify the expression by combining like terms. Remember, like terms have the same variable part, to simplify these terms, one performs operations between the coefficients. Please note that a variable with an exponent is not the same as a variable without the exponent. A term with no variable part is referred to as a constant, constants are like terms.

$2h^2-7h+2h^2-h+6+4h-9h$

$(2h^2+2h^2)+(-7h-h+4h-9h)+(6)$

$4h^2-13h+6$

2.

Use a very similar method to solve this problem as used in the first. Please note that all of the rules mentioned in the first problem also apply to this problem; for that matter, the rules mentioned in the first problem generally apply to any pre-algebra problem.

$8x^3y^2-7x^2y+8x-4-x^3y^2+2x^2y+4x^2y-8x+5$

$(8x^3y^2-x^3y^2)+(-7x^2y+2x^2y+4x^2y)+(8x-8x)+(-4+5)$

$7x^3y^2-x^2y+1$

3.

Use the same rules as applied in the first problem. Also, keep the distributive property in mind. In simple terms, the distributive property states the following ( $a(b+c)=(a)(b)+(a)(c)=ab+ac$ ). Also note, a term raised to an exponent is equal to the term times itself the number of times the exponent indicates. In the event that the term raised to an exponent is a constant, one can simplify it. Apply these properties here,