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vlabodo [156]
3 years ago
10

Evaluate function expressions 3•f(-4)- 3•g(-2)

Mathematics
2 answers:
Tanya [424]3 years ago
7 0

Answer:

-12f + 6g

Step-by-step explanation:

f x -4 then 3 x -4f which you get -12f.

g x -2= -2g then -2g x 3= -6g.

Then we take the subtraction sign in the middle and make the equation simpler.

MAXImum [283]3 years ago
6 0

Answer:3(-4f-g(to the power of -2)

Step-by-step explanation:

3(-fx4-g(to the power of -2) reorder and get solution.

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How do you solve these
IRINA_888 [86]
In the first equation, you can tell that the x's are being multiplied by 3 to get to the y's so the blank will be 9. The second table would not be a function since all the x values are not being multiplied by the same number to get the y value. Hope this helps!

4 0
3 years ago
Factor completely<br> m^2 — Зm — 18
lilavasa [31]

Answer:

(m + 3) (m - 6)

Step-by-step explanation:

Simplify the following:

m^2 - 3 m - 18

The factors of -18 that sum to -3 are 3 and -6. So, m^2 - 3 m - 18 = (m + 3) (m - 6):

Answer: |

| (m + 3) (m - 6)

4 0
3 years ago
These tables represent continuous functions. In which table do the values represent an exponential function.
Dennis_Churaev [7]
If you look at all 4 of these tables. Exponential means a large increase or gradual increase so. With that in mind if you look at table 1, the y column only increases by 5. For B, the y column only increases by 1. For C, the y column increase by 4 then 8 then 16 then 32. Therefore it is exponential. For D it is consistently an increase of 8.

Hope this helps!
4 0
3 years ago
Read 2 more answers
Does the table represent a proportional relationship? Show your work!
wel
Yes, adding going up one every time, 5,6,7,8
4 0
3 years ago
If a+b+c=0 then find the value of a^2+b^2+c^2/a^2-bc pls help me
VladimirAG [237]

a+b+c=0

[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]

[a^2+b^2+c^2+2ab+2ac+2bc=0]

[a^2+b^2+c^2=-(2ab+2ac+2bc)]

[a^2+b^2+c^2=-2(ab+ac+bc)] (i)

also

[a=-b-c]

[a^2=-ab-ac] (ii)

[-c=a+b]

[-bc=ab+b^2] (iii)

adding (ii) and (iii) ,we have

[a^2-bc=b^2-ac] (iv)

devide (i) by (iv)

[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
8 0
3 years ago
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