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Nookie1986 [14]
3 years ago
6

Which statements are true for the functions g(x) = x2 and h(x) = –x2 ? Check all that apply.

Mathematics
2 answers:
mrs_skeptik [129]3 years ago
4 0

Answer:

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

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

Step-by-step explanation:

For any value of x g(x) is always greater than h(x) and for any value of x, h(x) will always be greater than g(x) are not true.

The given function is:

g(x) = x^2 and h(x) = –x^2

x=0

g(0)=(0)^2 = 0

h(0)= -(0)^2 = 0

Now check the condition for x = -1

put x =-1 in the given functions.

g(x)=x^2

g(-1) = (-1)^2 = 1

h(x)= -x^2

h(-1) =  -(-1)^2 = -1

g(x)>h(x)

Now take a positive value of x= 3

Put the value in the given functions:

g(3) = (3)^2 = 9

h(3) = -(3)^2 = -9

g(x)>h(x)

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

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

fomenos3 years ago
4 0

Answer:

Simplified Answer: C & F

Step-by-step explanation:

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The answer is 1000000
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in a classroom, 1/6 of the students are wearing blue shirts and 2/3 are wearing white shirts. there are 36 students in the class
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The fraction indicating the number of students who are not wearing either blue or white shirt is:
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Identify the independent variable and the dependent variable.
melamori03 [73]

Answer:

no

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
PLEASE HELP Word problem involving average rate of change
oee [108]

Answer:

(a): 26.1

(b): 27.9

Please see below for the steps.

Step-by-step explanation:

(a):

Use points (0,0) and (3, 78.3)

Use slope formula. The slope formula is also used to find average rate of change (just so you know).

y2-y1/x2-x1

78.3-0/3-0=78.3/3=26.1

Answer for (a) is 26.1

(b):

Use points (4, 147.6) and (9, 287.1)

Use slope formula.

y2-y1/x2-x1

287.1-147.6/9-4=139.5/5=27.9

The answer for (b) is 27.9

Hope this helps!

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Have a great day!

5 0
1 year ago
A new centrifugal pump is being considered for an application involving the pumping of ammonia. The specification is that the fl
ollegr [7]

Answer:

from the t-distribution table, at df = 7 and t = 2.23

Lies p-values [ 0.05 and 0.025 ]

Hence;

0.025 < p-value < 0.05

Step-by-step explanation:

Given that;

x^{bar} = 6.5 gpm

μ = 5 gpm

n = eight runs = 8

standard deviation σ = 1.9 gpm

Test statistics;

t = (x^{bar} - μ) / \frac{s}{\sqrt{n} }

we substitute

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t = 1.5 / 0.67175

t = 2.23

the degree of freedom df = n-1 = 8 - 1

df = 7

Now, from the t-distribution table, at df = 7 and t = 2.23

Lies p-values [ 0.05 and 0.025 ]

Hence;

0.025 < p-value < 0.05

3 0
3 years ago
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