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irga5000 [103]
3 years ago
13

Rewrite the expression (4x^2+5x)^2-5 (4x^2+5x)-6 as a product of four linear factors.

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0
=<span><span><span><span><span>16<span>x4</span></span>+<span>40<span>x3</span></span></span>+<span>5<span>x2</span></span></span>−<span>25x</span></span>−<span>6</span></span>
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Simple, it is equal to |4|, you have to draw the lines around the 4 on the worksheet.
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Just checking my answers guys! what is 8x20
marissa [1.9K]
160 is the answer to 8x20
3 0
3 years ago
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1) Write a system of equations to describe the situation below, solve using any method, and
Aloiza [94]
This situation can be represented by

y = 33x + 223

y = 32x + 239

I still like substitution, so let’s plug in 33x + 223 into the second equation

33x + 223 = 32x + 239

Move the terms into appropriate sides

33x - 32x = 239 - 223

Combine like terms

x = 16

Then plug in x for any equation

y = 32(16) + 239

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7 0
3 years ago
At the beginning of a population study, the population of a large city was 1.65 million people. Three years later, the populatio
Maksim231197 [3]

Answer:

C. 1.8027

Step-by-step explanation:

The exponential population growth model is given by:

P(t) = P_{0}e^{rt}

In which P(t) is the population after t years, P_{0} is the initial population and r is the growth rate.

At the beginning of a population study, the population of a large city was 1.65 million people. Three years later, the population was 1.74 million people.

This means that P_{0} = 1.65, P(3) = 1.74

Applying this to the equation, we find r. So

P(t) = P_{0}e^{rt}

1.74 = 1.65e^{3r}

e^{3r} = \frac{1.74}{1.65}

e^{3r} = 1.0545

Applying ln to both sides

\ln{e^{3r}} = \ln{1.0545}

3r = \ln{1.0545}

r = \frac{\ln{1.0545}}{3}

r = 0.0177

So

P(t) = 1.65e^{0.0177t}

What would be the population of the city 5 years after the start of the population study?

This is P(5).

P(t) = 1.65e^{0.0177t}

P(5) = 1.65e^{0.0177*5}

P(5) = 1.8027

So the correct answer is:

C. 1.8027

6 0
3 years ago
Use the quadratic formula to find the solutions to the quadratic equation below. Check all that apply. 5x2-X-4 = 0 A. -4/5 B. 5/
Alenkasestr [34]

Hi

5x²-x-4 = 0  

Δ= (-1)² - 4*5*(-4)

Δ =  1 -4*-20

Δ = 1  +80

Δ = 81  

√Δ=  9

as Δ ≥ 0 , so 2 solutions exist in R

S1 is :    ( 1+9) /2*5  =   10/10 = 1

s2 =      (1 -9)/2*5 =  -8/10 = -4*2 /2*5  = -4/5

Corrects answers are  A  and  D

6 0
3 years ago
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