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xz_007 [3.2K]
3 years ago
13

Help!I got stuck and don't know what to do.

Mathematics
1 answer:
lutik1710 [3]3 years ago
7 0
P/6
j-65
185+h
16g
4x
8•9=72
11+7=18
4(2+7)=36
60/5=12
6•8÷2=24
1 to 3 2 to 1 3 to 2
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Suppose f(x)=x^2 what is the graph of g(x)= f(5x)
Novosadov [1.4K]

Answer:

Step-by-step explanation:

That '5' in f(5x) will compress the graph of x^2 horizontally.

If you were to graph f(x) = x^2, you'd get a parabolic graph; the parabola will open UP.

Suppose you graphed f(x) = x^2 on the interval [-4, 4].

Then the graph of g(x) = f(5x) would be graphed on the interval [-4/5, 4/5].  In other words, the graph would be on a shorter interval, one shorter by a factor of 5.

6 0
3 years ago
Read 2 more answers
Can you help me find the missing value to the nearest hundredth?! <br>cos□= 2/5 <br><br>Thank you ❤
Aleksandr [31]

Answer:

66.42°

Step-by-step explanation:

Given

cos ? = \frac{2}{5} , then

? = cos^{-1} (\frac{2}{5} ) ≈ 66.42° ( to the nearest hundredth )

7 0
3 years ago
What is the unit rate for the table showing the cost of movies?
photoshop1234 [79]
2.75! Good luck hope you do good
6 0
3 years ago
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Is the point (4,2) the solution to the system? Why or why not? Explain your reasoning.
VMariaS [17]
Solve the top equation to get on variable on one side
2x-y=6
-y=-2x+6
y=2x-6
then plug in the coordinate for x and y for both equations. if both equations are true then it’s a solution
2=2(2)+6
2=10

4+2=6

so it’s not a solution
4 0
3 years ago
A study of 25 graduates of 4-year public colleges revealed the mean amount owed by a student in student loans was $55,051. The s
Harman [31]

Answer:

Step 1

The data represent amount.

A 90% confidence interval for the population mean is,

First, compute t-critical value then find confidence interval.

The t critical value for the 90% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed and the row headed in the t distribution table by using degree of freedom is,

The t critical value for the 90% confidence interval is 1.711.

A 90% confidence interval for the population mean is .

Step 2

It is reasonable to conclude that mean of the population is actually $55000 due to a 90% confidence intrerval for population mean is between $52461.23 and $57640.77 does include $55000.

The data represent amount.

A 90% confidence interval for the population mean is,

First, compute t-critical value then find confidence interval.

The t critical value for the 90% confidence interval is,

The sample size is small and two-tailed test. Look in the column headed and the row headed in the t distribution table by using degree of freedom is,

The t critical value for the 90% confidence interval is 1.711.

A 90% confidence interval for the population mean is .

It is reasonable to conclude that mean of the population is actually $55000 due to a 90% confidence intrerval for population mean is between $52461.23 and $57640.77 does include $55000.

5 0
2 years ago
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