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never [62]
2 years ago
12

Which statements are true for the functions g(x)=x^2 and h(x)=-x^2? Check all that apply

Mathematics
1 answer:
galina1969 [7]2 years ago
4 0

Answer:

if x=0 then they have same value

1 and 2 options are out

for x=-1

g(-1)=1

h(-1)=-1

3 is true

4th

FALSE

for all values except 0, g(x)>h(x)

correct ones are

g(x) > h(x) for x = -1.

For positive values of x, g(x) > h(x).

For negative values of x, g(x) > h(x).

Step-by-step explanation:

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Sara lost 12 pounds over the summer by jogging each week. By winter time she had gained 4 3/8 pounds. Represent this situation w
Vika [28.1K]

The situation with an expression involving signed numbers is w - 12 + 4 3/8 and the overall change in Sara’s weight is -7 5/8 pounds

<h3>Represent this situation with an expression involving signed numbers. </h3>

Let Sara's initial weight be w.

The given parameters are:

Lost = 12 pounds

Gain = 4 3/8 pounds

The situation with an expression involving signed numbers is represented as:

Weight = Initial - Lost + Gain

So, we have:

Weight = w - 12 + 4 3/8

<h3>What is the overall change in Sara’s weight?.</h3>

In (a), we have:

Weight = w - 12 + 4 3/8

The overall change is:

Change = - 12 + 4 3/8

Evaluate

Change = -7 5/8

Hence, the overall change in Sara’s weight is -7 5/8 pounds

Read more about sum and difference at:

brainly.com/question/17695139

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3 0
1 year ago
Julie bought a drink for $2.99, two bags of chips for $1.29 each, and some gasoline at a convenience store. The price of gasolin
mylen [45]
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7 0
2 years ago
What is the value of x? x+15=2x-40
IrinaVladis [17]

Answer:

55 =x

Step-by-step explanation:

x+15=2x-40

Subtract x from each side

x-x+15=2x-x-40

15 = x-40

Add 40 to each side

15+40 = x-40+40

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7 0
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What is the average speed if Time = 3 Distance = 90
Natasha2012 [34]
Speed= distance divided by time. The answer is 30, and im assuming miles per hour. Hope this helps
6 0
3 years ago
The question is in the picture​
Fudgin [204]

Answer:

(x) = 1/2x-3/2

Step-by-step explanation:

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3 years ago
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