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Semenov [28]
3 years ago
15

Evaluate 15/r -1 when r = 5

Mathematics
2 answers:
Juli2301 [7.4K]3 years ago
4 0
\frac{15}{r}-1=\frac{15}{5}-1=3-1=2

2)
\frac{15}{r-1}=\frac{15}{5-1}=\frac{15}{4}=3\frac{3}{4}=3.75
Fynjy0 [20]3 years ago
4 0
15/r-1  
15/5-1  (substitute 5 for r)
15/4     (5-1 is 4)
3.75     (15/4 is 3.75)
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Find a polynomial equation that has zeros at x = -4, x = 0, and a double root at x = 7.
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D

Step-by-step explanation:

Given a polynomial f(x) with roots say x = a and x = b, then the factors are

(x - a) and (x - b) and f(x) is the product of the roots, that is

f(x) = (x - a)(x - b)

Here the roots are x = - 4, x = 0 and double root at x = 7

Thus factors are (x - (- 4)), (x - 0), that is (x + 4) and x

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3 years ago
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fredd [130]

Answer:

\frac{4k + 2}{k^{2}-4 }  ×  \frac{k-2}{2k+1}   =  \frac{2}{k + 2}

Step-by-step explanation:

\frac{4k + 2}{k^{2}-4 }  ×  \frac{k-2}{2k+1}

To solve the above, we need to follow the steps below;

4k+2 can be factorize, so that;

4k +2 = 2 (2k + 1)

k² - 4  can also be be expanded, so that;

k² - 4 = (k-2)(k+2)

Lets replace  4k +2  by  2 (2k + 1)

and

k² - 4 by  (k-2)(k+2)   in the expression  given

\frac{4k + 2}{k^{2}-4 }  ×  \frac{k-2}{2k+1}

\frac{2(2k+ 1)}{(k-2)(k+2)}   ×  \frac{k-2}{2k+1}

(2k+1) at the numerator will cancel-out (2k+1) at the denominator, also (k-2) at the numerator will cancel-out (k-2) at the denominator,

So our expression becomes;  

\frac{2}{k + 2}

Therefore, \frac{4k + 2}{k^{2}-4 }  ×  \frac{k-2}{2k+1}   =  \frac{2}{k + 2}

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3 years ago
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Step-by-step explanation:

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icang [17]

Answer:

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Step-by-step explanation:

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