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MAXImum [283]
4 years ago
13

Solve by using tables. Give each answer to at most two decimal places. –2x2 – 4 = –8x (1 point) 0.59, 3.41 –3.41, –0.59 1.17, 6.

83 –1.41, 1.41
Mathematics
2 answers:
Paul [167]4 years ago
7 0
The answer is ou need to rewrite the equation to make it easy... and let it equal zero <span>2<span>x2</span>−8x+4=0</span> which will become <span>2(<span>x2</span>−4x+2)=0</span> you need to solve <span><span>x2</span>−4x+2=0</span><span> by completing the square or the general quadratic formula.</span>
slamgirl [31]4 years ago
4 0

Answer:

The solution to the quadratic equation x² - 4x - 2 =0 is 0.59 and 3.41

Step-by-step explanation:

Given the equation -2x² - 4 = -8x

rearrange the equation we now have -2x² +8x - 4 = 0

divide each term in the equation by -2, we have

x² - 4x - 2 =0

To get the table, we substitute the given values of x in the equation to get the corresponding values for y

x values     y values

0.59            -0.0119

3.41              -0.0119

-3.41              27.2681

-0.59            4.7081

1.17                -1.3111

6.83             21.3289

-1.41              9.6281

1.41               -1.6519

The two points of x where y =0 are 0.59 and 3.41

Therefore the solution to the quadratic equation x² - 4x - 2 =0 is 0.59 and 3.41

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Answer:

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2 years ago
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Gala2k [10]

Answer:

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Step-by-step explanation:

7 = -2*-3 + 1 makes sense

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6 0
3 years ago
What is the value of x, when 10(x + 2) = 5(x + 8)?
olga_2 [115]
Solve for x:
10 (x + 2) = 5 (x + 8)

Expand out terms of the left hand side:
10 x + 20 = 5 (x + 8)

Expand out terms of the right hand side:
10 x + 20 = 5 x + 40

Subtract 5 x from both sides:
(10 x - 5 x) + 20 = (5 x - 5 x) + 40

10 x - 5 x = 5 x:
5 x + 20 = (5 x - 5 x) + 40

5 x - 5 x = 0:
5 x + 20 = 40

Subtract 20 from both sides:
5 x + (20 - 20) = 40 - 20

20 - 20 = 0:
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40 - 20 = 20:
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Divide both sides of 5 x = 20 by 5:
(5 x)/5 = 20/5

5/5 = 1:
x = 20/5

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6 0
4 years ago
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3 years ago
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poizon [28]

Answer:

<em>m<A = 40 deg</em>

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Step-by-step explanation:

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6 0
4 years ago
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