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ser-zykov [4K]
3 years ago
10

the sum of −2x2 + x + 31 and 3x2 + 7x − 8 can be written in the form ax2 + bx + c, where a, b, and c are constants. What is the

value of a + b + c ?
Mathematics
1 answer:
just olya [345]3 years ago
8 0

Answer:

Part 1) a=1,b=8,c=23

part 2) a+b+c=32

Step-by-step explanation:

we have

(-2x^{2} +x+31)+(3x^{2}+7x-8)

Adds the terms

Group terms that contain the same variable

(-2x^{2}+3x^{2})+(x+7x)+(31-8)

Combine like terms

(x^{2})+(8x)+(23)

where

a=1,b=8,c=23

therefore

a+b+c=1+8+23=32

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Find the perimeter. Simplify your answer completely.
Vladimir79 [104]

Answer:119 4/7

Step-by-step explanation:

2/14+6/14+2/14+4/14+1/14+3/14+4/14= 22/14

10+35+10+12+24+15+12=118

118 22/14

119 8/14

119 4/7

5 0
3 years ago
Read 2 more answers
Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.
Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ;  x = \frac{2-(3)\sqrt{2}}{2}

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the  following condition must be followed:

Either  (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}

for the equation ax² + bx + c = 0

thus,

the roots are

x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}

or

x = \frac{6\pm\sqrt{36-12}}{2}

or

x = \frac{6+\sqrt{24}}{2} and, x = x = \frac{6-\sqrt{24}}{2}

or

x = \frac{6+2\sqrt{6}}{2} and, x = x = \frac{6-2\sqrt{6}}{2}

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}

or

x = \frac{4\pm\sqrt{16+56}}{4}

or

x = \frac{4+\sqrt{72}}{4} and, x = x = \frac{4-\sqrt{72}}{4}

or

x = \frac{4+\sqrt{2^2\times3^2\times2}}{2} and, x = x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}

or

x = \frac{4+(2\times3)\sqrt{2}}{2} and, x = x = \frac{4-(2\times3)\sqrt{2}}{4}

or

x = \frac{2+3\sqrt{2}}{2} and, x = \frac{2-(3)\sqrt{2}}{2}

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; x = \frac{2+3\sqrt{2}}{2} ; x = \frac{2-(3)\sqrt{2}}{2}

7 0
3 years ago
Pls help me!!! ( pls dont answer if you don’t have the answer I will report you ...).
bezimeni [28]
I think 517.26
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5 0
3 years ago
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What is the range of the function on the graph?
Ainat [17]
Check the picture below.

includes 2, notice, it doesn't go below 2 over the y-axis, and then it keeps on going up.  So, the range is from +2 onwards.

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4 years ago
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0.0660 into scientific notation
Yakvenalex [24]

Answer:

0.0660 = 6.60 x 10-2

Step-by-step explanation:

5 0
3 years ago
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