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3 years ago

12

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! Simplify.

2 answers:

uranmaximum [27]3 years ago

8 0

Answer: $\bold{B)\ \dfrac{x-2}{x+1}}$

Step-by-step explanation:

Factor the numerator (top of fraction) and denominator (bottom of fraction). Then cancel out the common factor.

To factor, find two numbers whose product is the c-value and sum is the b-value.

x² + 5x - 14

∧

-1 +14 = 13

-2 +7 = 5 this works!

So, x² + 5x - 14 = (x - 2)(x + 7)

x² + 8x + 7

∧

1 +7 = 8 this work!

So, x² + 8x + 7 = (x + 1)(x + 7)

$\dfrac{x^2+5x-14}{x^2+8x+7}=\dfrac{(x+7)(x-2)}{(x+7)(x+1)}=\dfrac{x-2}{x+1}$

nikdorinn [45]3 years ago

4 0

X^2+5x-14 = (x+7)(x-2)
X^2+8x+7=(x+7)(x+1)

(X+7) crosses out

Answer x-2 /x+1

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What is 15% of 60? create a model to prove your answer. in equivalent fractions

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2 years ago

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Bezzdna [24]

Since $a_1,a_2,a_3,\cdots$ are in arithmetic progression,

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$a_4 = a_3+2 = a_1+3\cdot2$

$\cdots \implies a_n = a_1 + 2(n-1)$

and since $b_1,b_2,b_3,\cdots$ are in geometric progression,

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$b_4=2b_3=2^3b_1$

$\cdots\implies b_n=2^{n-1}b_1$

Recall that

$\displaystyle \sum_{k=1}^n 1 = \underbrace{1+1+1+\cdots+1}_{n\,\rm times} = n$

$\displaystyle \sum_{k=1}^n k = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}2$

It follows that

$a_1 + a_2 + \cdots + a_n = \displaystyle \sum_{k=1}^n (a_1 + 2(k-1)) \\\\ ~~~~~~~~ = a_1 \sum_{k=1}^n 1 + 2 \sum_{k=1}^n (k-1) \\\\ ~~~~~~~~ = a_1 n + n(n-1)$

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$\displaystyle \sum_{k=1}^n ar^{k-1} = \frac{a(1-r^n)}{1-r}$

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$2^n > \dfrac4{\ln(2)} n \implies \ln(2)\cdot2^n - 2 > 4n - 2$

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1 year ago

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