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4 years ago

Helpppppppppppppoppppp

Mathematics

1 answer:

STALIN [3.7K]4 years ago

3 0

Answer:

it would be 75

Step-by-step explanation:

if add 10+ 10+ 30 + 25 its would be 75

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Y = -3x + 3 HELP I WILL GIVE BRAINLIEST TO BEST ANSWER 40 POINTS

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Answer:

x=anything

Step-by-step explanation:

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Are s^2 ,2s^2 and 3s^2 equivalent

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No they are not quivalent

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3 years ago

Kendra earns $70 per week mowing lawns. She saves 60% of the money she earns each week. How many weeks will it take Kendra to sa

katrin [286]

Answer:

5 weeks

Step-by-step explanation:

Let's find how much money she saves a week by multiplying the decimal form of 60% and $70.

To get the decimal form of a percent you have to move the decimal to the left twice, like this: 60 ---> .60

Now that we have the decimal form of the percent we can multiply it by the 70 to find how much she saves a week:

(70)(.6) = 42

Therefore, Kendra saves $42 each week.

To find how many weeks it will take Kendra to buy the game system you divide $210 by the $42. Like this:

(210)/(42) = 5

All in all, it will take Kendra 5 weeks to save enough money to buy a game system that costs $210.

I hope this helps!!

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3 years ago

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3 years ago

Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

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

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

case B) $f(x)=x(x+2)$

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

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

case C) $f(x)=(x-2)(x+2)$

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

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

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

case D) $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

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

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

the answer is

$f(x)=(x-2)(x+2)$

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3 years ago

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