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Olenka [21]
3 years ago
10

Question 17 PLEASE HELP given the function: f(x) =3x+7. Find the value of y when x=-8

Mathematics
1 answer:
STALIN [3.7K]3 years ago
8 0

→Answer:

y = -17

→Step-by-step explanation:

Well first off the f(x) in the equation f(x) = 3x + 7 can also be written as y,

y = 3x + 7

So if x is -8 we need to plug it in for x and solve, simple as that :)

y = 3(-8) + 7

y = -24 +7

y = -17

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andrey2020 [161]

Answer:

9,999,995

Step-by-step explanation:

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How to solve function operations
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3 0
3 years ago
FIND NTH TERM QUICKEST IN GETS BRAINLIEST
patriot [66]
Answer is 13 because the numbers are increasing though odd numbers that are also increasing if that makes sense.

It’s going from 4 > 9 > 16 > 25 > 36

so from 4-9 it adds 5 and from 9-16 it adds 7 and goes on so that’s it’s like

+5 > +7 > +9 > +11

sorry if it doesn’t make good sense
4 0
3 years ago
A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee
katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

^nC_r=\dfrac{n!}{(n-r)!r!}

Given : Total multiple-choice questions  = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= ^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84

No. of ways to answer 4 open-ended problems

= ^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= 84\times15=1260

Hence, the correct answer is option A) 1260

5 0
4 years ago
What is x+y=10;y=x+10
Alenkasestr [34]

[ Answer ]

\boxed{\bold{X \ = \ 0, \ Y \ = \ 10}}

[ Explanation ]

  • System Of Equations \bold{\left \{ {X \ + \ Y \ = \ 10} \atop {Y \ + \ X \ = \ 10}} \right.}

-----------------------------------

  • [Substitute] Y = X + 10

\bold{\begin{bmatrix}x+x+10=10\end{bmatrix}}

  • Isolate x for x + x + 10 = 10: x = 0

For y = 10

Substitute x = 0

Y = 0 + 10

0 + 10 = 10

y = 10

  • Solutions

X = 0

Y = 10

\boxed{\bold{[] \ Eclipsed \ []}}

3 0
3 years ago
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