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Dvinal [7]
3 years ago
14

Find the zeros of the polynomial 12x^2+31x+20

Mathematics
1 answer:
alexdok [17]3 years ago
8 0

Answer:

-5/4; -4/3

Step-by-step explanation:

12x^2 + 31x + 20 =

(4x + 5)(3x + 4) -->

x = -5/4 OR -4/3

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Find the value of the missing coefficient a in the equation y = a(x - 3)(x + 5) if the graph goes through the point (1, -6)
horrorfan [7]

Answer:

3

Step-by-step explanation:

you have to solve for x then you can find the coeeffiecnt

3 0
2 years ago
Use synthetic substitution to find f (-5) for f (x) = x3 – 5x +2.
Sonbull [250]

Answer:

_5(_5)=25

25+2=27

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8 0
3 years ago
Read 2 more answers
B + 12 = 16; 2, 3, 4
Molodets [167]

Answer:

4

Step-by-step explanation:

b+12 =16

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hope this helps

5 0
3 years ago
Please help thank you
Keith_Richards [23]

Answer:

5

alternates

Step-by-step explanation:

The series is for k = 0 to 4, so k = 0, 1, 2, 3, 4.

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A negative number raised to an odd integer is negative.

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7 0
3 years ago
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pashok25 [27]

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