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2 years ago

What is the factored form of this expression? x2 + 15x + 56

Mathematics

2 answers:

Evgen [1.6K]2 years ago

8 0

Answer:

(x+8)(x+7)

Step-by-step explanation:

AlladinOne [14]2 years ago

6 0

Step-by-step explanation:

${x}^{2} + 15x + 56$

${x}^{2} + 7x + 8x + 56$

taking common

$x(x + 7) + 8(x + 7)$

$(x + 8) (x + 7)$

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The difference of eighteen and a number is 3

Firdavs [7]

Answer:

Step-by-step explanation:

18-x=3

18-3=x

18-3=15

5 0

2 years ago

The projected rate of increase in enrollment at a new branch of the UT-system is estimated by E ′ (t) = 12000(t + 9)−3/2 where E

nexus9112 [7]

Answer:

The projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

Step-by-step explanation:

Consider the provided projected rate.

$E'(t) = 12000(t + 9)^{\frac{-3}{2}}$

Integrate the above function.

$E(t) =\int 12000(t + 9)^{\frac{-3}{2}}dt$

$E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+c$

The initial enrollment is 2000, that means at t=0 the value of E(t)=2000.

$2000=-\frac{24000}{\left(0+9\right)^{\frac{1}{2}}}+c$

$2000=-\frac{24000}{3}+c$

$2000=-8000+c$

$c=10,000$

Therefore, $E(t) =-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

Now we need to find $\lim_{t \to \infty} E(t)$

$\lim_{t \to \infty} E(t)=-\frac{24000}{\left(t+9\right)^{\frac{1}{2}}}+10,000$

$\lim_{t \to \infty} E(t)=10,000$

Hence, the projected enrollment is $\lim_{t \to \infty} E(t)=10,000$

8 0

2 years ago

Given the function f(x) = 6x2 + 3 Find f(-2) -22 27 9 15

Andrews [41]

Answer:

Step-by-step explanation:

Out of those numbers and assuming the equation is f(x) = 6x^2=3

f(-2)= 6(-2)^2+3

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2 years ago

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nevsk [136]

Answer:

Quadrant IV is always in the bottom right corner.

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2 years ago

If h(c) = -4, then c =

klemol [59]

Answer should be -4

since there is no other context, we can assume that x is -4.

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3 years ago