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Sonbull [250]
3 years ago
6

Which polynomial function could be represented by the graph below?

Mathematics
2 answers:
miss Akunina [59]3 years ago
6 0

Answer:

f(x) = x² - 6x + 8

Step-by-step explanation:

mr_godi [17]3 years ago
4 0

f(x) = x² - 6x + 8

from the graph the zeros are x = 2 and x = 4, hence

the factors are (x - 2) and (x - 4), thus

f(x) = (x - 2)(x - 4) ← expand factors

f(x) = x² - 6x + 8 is a possible polynomial


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WILL GIVE BRAINLET !!!<br> 2 1/12 + 4/12 ( show how u her the answer )
spin [16.1K]

Answer:

2 5/12

Step-by-step explanation:

you can combine two fractions if they have the same denominator (bottom number) by adding the numerators (top numbers) and keeping the bottom number the same.

1/12+4/12 is the same as 1+4 you just put the sum over 12:

2 1/12+4/12= 2 5/12

it can also be written as 29/12

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3 years ago
Need help with this please
Katarina [22]

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3 years ago
How many times does 32 go into 82
pav-90 [236]
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3 years ago
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vovangra [49]

Answer:

1. 27^{\frac{2}{3} } =9

2. \sqrt{36^{3} } =216

3. (-243)^{\frac{3}{5} } =-27

4. 40^{\frac{2}{3}}=4\sqrt[3]{25} =4325

5. Step 4: (\frac{343}{27}) ^{-1} =\frac{27}{343}

6. D. -72cd^{7}

Step-by-step explanation:

Use the following properties:

a^{\frac{x}{y} } =\sqrt[x]{a^{y} }

\sqrt[n]{ab} =\sqrt[n]{a} \sqrt[n]{b}

a^{-n} =\frac{1}{a^{n} }

(xy)^{z} =x^{z} y^{z} \\\\

(x^{y}) ^{z} =x^{yz}

x^{y} x^{z} =x^{y+z}

So:

1. 27^{\frac{2}{3} } =\sqrt[3]{27^{2}} =\sqrt[3]{729} }=9

2. \sqrt{36^{3} } =\sqrt{36*36*36} =\sqrt{36} \sqrt{36}  \sqrt{36} =6*6*6=216

3. (-243)^{\frac{3}{5} } =\sqrt[5]{-243^{3} } =\sqrt[5]{-14348907} =-27

4. 40^{\frac{2}{3}}=\sqrt[3]{40^{2} } =\sqrt[3]{2^{6} 5^{2} } =\sqrt[3]{2^{6} } \sqrt[3]{5^{2} } =2^{\frac{6}{3} } 5^{\frac{2}{3} } =4 *5^{\frac{2}{3} } =4\sqrt[3]{5^{2} } =4\sqrt[3]{25}=4325

5. (\frac{343}{27}) ^{-1} =\frac{1}{\frac{343}{27} } =\frac{27}{343}

6.

(-8c^{9} d^{-3} )^{\frac{1}{3} } *(6c^{-1}d^{4})^{2} =\sqrt[3]{-8} c^{3} d^{-1} 36c^{-2} d^{8} \\\\-2c^{3} d^{-1} 36c^{-2} d^{8}=-72cd^{7}

7 0
3 years ago
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