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Tresset [83]
3 years ago
10

Question 19 What must be done to the graph of f(x) = |x| to obtain the graph of the function g(x) = 3/7 |x+4|-10?

Mathematics
1 answer:
Sergio [31]3 years ago
5 0

Answer:

The graph of f is shifted left 4 units, vertically shrunk by a factor of \frac{3}{7} and shifted down 10 units.

Step-by-step explanation:

The base graph is f(x)=|x|

The transformed graph is g(x)=\frac{3}{7}|x+4|-10

Let us rewrite g(x) in terms of f(x) to get:

g(x)=\frac{3}{7}f(x+4)-10

This means that f(x) was vertically shrunk by a factor of \frac{3}{7} and shifted  4 units to the left and 10 units down.

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I need help simplifying 5 1/2 ÷ 2 1/2 then writing it as a mixed number ​
Yakvenalex [24]

Answer:

exact form

11/5

decimal form 2.2

mixed number form 2 1/5

Step-by-step explanation:

6 0
2 years ago
Rewrite this polynomial in STANDARD FORM:<br>5x-10x^2+8x^3​
Ainat [17]

Answer:

8 {x}^{3}   -  10 {x}^{2}  + 5x

Step-by-step explanation:

5x - 10 {x}^{2} + 8 {x}^{3}   \\   \red{ \bold{= 8 {x}^{3}   -  10 {x}^{2}  + 5x}} \\ is \: in \: the \: standard \: form \\

4 0
3 years ago
The population of a town can be modeled by P= 22,000 + 125t, where P represents the population and the t represents the number o
nlexa [21]
Given that the population can be modeled by P=22000+125t, to get the number of years after which the population will be 26000, we proceed as follows:
P=26000
substituting this in the model we get:
26000=22000+125t
solving for t we get:
t=4000/125
t=32
therefore t=32 years
This means it will take 32 years for the population to be 32 years. Thus the year in the year 2032
7 0
3 years ago
Six friends attend a party. They form pairs for a game. How many different pairs are possible?
Leona [35]

Pairs, in this case, relates to a group of 2 or more. We have 6 friends. Let's call them A,B,C,D,E,F. This will allow us to make a [some sort of] combination tree:

1. ABC against DEF

2. ABD against CEF

3. ABE against CDF

4. ABF against CDE

5. ACD against BFE

6. ACE against BDF

7. ACF against BDE

8. ADE against BCF

9. ADF against BCE

10. AEF against BCD

I believe there are 12 combinations... I just can't think of the last 2 though.

5 0
3 years ago
2-6=? need help please
suter [353]
The answer is -4 hope that helps :)
5 0
3 years ago
Read 2 more answers
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