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Snezhnost [94]
3 years ago
12

Please help. Write a situation that can be represented with this graph. ​

Mathematics
1 answer:
Rzqust [24]3 years ago
6 0
Ion know sorry buddy I wish I could help
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Solve the equation <br> 3(x-8)=15<br> x=___
KengaRu [80]
X = 13
Hope that helps (: 
If you need me to show my work letme know!
6 0
3 years ago
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I GIVE BRAINLIEST :)<br><br> Question is in picture, tysm
77julia77 [94]

Answer:

No x^3 times x^5 is not equal to x^15. It is equal to x^8

Step-by-step explanation:

x3x5

=x3*x5

=(x*x*x)*(x*x*x*x*x)

=x*x*x*x*x*x*x*x

=x8

7 0
3 years ago
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Jalen bought 24 juice packs for $7.20. Tia bought 6 juice packs for $2.70. Which statement describes the difference in the unit
Wittaler [7]
The operation is times then subtract because when the ? says difference that mean subtract and it's times because it says 6 juice packs for $2.70 and that's the operations
4 0
4 years ago
What is the value of StartFraction negative 8 (17 minus 12) over Negative 2 (8 minus (negative 2)) EndFraction?
Ahat [919]

Answer:

2

Step-by-step explanation:

\frac{-8(17-12)}{-2(8-(-2))} \\=\frac{-8(5)}{-2(8+2)} \\=\frac{-40}{-2(10)} \\=\frac{-40}{-20} \\=2

6 0
3 years ago
Which function has an asymptote at x= 39
LekaFEV [45]

Answer:

option B : f(x)= log_7(x-39)

Step-by-step explanation:

(a) f(x) = 7^{x-39}

For exponential function , there is no vertical asymptotes

General form of exponential function is

f(x) = n^{ax-b}+k

f(x) = 7^{x-39}

In the given f(x) the value of k =0

So horizontal asymptote is y=0

(b) lets check with option

f(x)= log_7(x-39)

To find vertical asymptote we set the argument of log =0  and solve for x

Argument of log is x-39

x-39=0 so x=39

Hence vertical asymptote at x=39


3 0
4 years ago
Read 2 more answers
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