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Andre45 [30]
3 years ago
8

Need help pls with this question. im struggling with this question.

Mathematics
1 answer:
jenyasd209 [6]3 years ago
5 0

Answer:

i can't rn but i could explain to you how... so first put a point on 0,5 then move down one and right one... then keep moving down one and right one

and there you go thats your graph

Step-by-step explanation:

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8th GRADE MATH
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Llana [10]

Answer:

  • See below

Step-by-step explanation:

#1 Synthetic method

  • 2x³ + x² - 2x - 4 ÷ x + 1 = 2x² - x - 1 rem 3

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x + 1 | 2x³ + x² - 2x + 4

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    -1 |  2       1    -2    4

       <u>|         - 2     1     1 </u>

          2     - 1    - 1    3

#2 Long division

  • 3x³ - 5x² - 26x - 8 ÷ 3x + 1 = x² - 2x - 8

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3x³ - 5x² - 26x - 8  <u>| 3x + 1</u>     = x² - 2x - 8

<u>3x³ + x²</u>

       -6x² - 26x

       <u>-6x² - 2x</u>

                  - 24x - 8

                  <u>- 24x - 8</u>

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8 0
3 years ago
165 is what percent of 433? set up a proportion and solve
seropon [69]

Answer:

38.1%

Step-by-step explanation:

Divide 165 by 433

165/433 = 0.381

Multiply the answer by 100

0.381 x 100 = 38.1

6 0
3 years ago
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Westkost [7]

Answer:

√2

Step-by-step explanation:

Angle ratio = 45 : 45 : 90

Ratio of side = 1 : 1 : √2

6 0
3 years ago
Verify a÷(b+c)not equal(a÷b)+a÷c for each of the following values of a, b and c
just olya [345]

Answer:

a) and b) satisfied the condition of the given equation.

Step-by-step explanation:

Given: \frac{a}{(b + c)} ≠ \frac{a}{b} + \frac{a}{c}

Then,

\frac{a}{(b + c)} ≠ \frac{(ac + ab)}{bc}

a) For a = 21, b = 4 and c = -2, then we have;

\frac{21}{(4 + (-2))} ≠ \frac{(21*-2 + 21*4)}{4*-2}

\frac{21}{4 - 2} ≠ \frac{(-42 + 84)}{-8}

\frac{21}{2} ≠ \frac{42}{-8}

\frac{21}{2} ≠ -\frac{21}{4}

Therefore the condition of the given equation is satisfied.

b) For a = 1, b = 10 and c = 1, then we have:

\frac{1}{(10 + 1)} ≠ \frac{(1*1 + 1*10)}{10*1}

\frac{1}{11} ≠ \frac{(1 + 10)}{10}

\frac{1}{11} ≠ \frac{11}{10}

Therefore the condition of the given equation is satisfied.

5 0
3 years ago
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