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erastovalidia [21]
3 years ago
6

Find the values for a and b that would make the equality true.

Mathematics
1 answer:
Crank3 years ago
6 0

Answer:

a = - 4, b = 5

Step-by-step explanation:

Expand the left side and compare the coefficients of like terms on the right side.

- 3(2x² + ax + b)

= - 6x² - 3ax - 3b

Comparing like terms with - 6x² + 12x - 15

x - term → - 3a = 12 ( divide both sides by - 3 )

                    a = - 4

constant term → - 3b = - 15 ( divide both sides by - 3 )

                               b = 5

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The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti
DiKsa [7]

Answer:

1) 20%

2) Choice a.

Step-by-step explanation:

P(t)=10000(0.2)^t

1) P(0) is the population initially.

P(1) is the population after a year.

\frac{P(1)}{P(0)} represents the population increase factor.

So let's evaluate that fraction:

\frac{P(1)}{P(0)}

\frac{10000(0.2)^1}{10000(0.2)^0}

\frac{0.2^1}{0.2^0}=\frac{0.2}{1}=0.2

0.2=20%

2) Let's figure out the population growth in terms of months instead of years.

P(t)=10000(0.2)^{t}

We want t to represent months.

A full year is 12 months, in a full year we have that P(1)=10000(0.2)^1=10000(0.2)=2000

So we want a new P such that P(12)=2000 since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) P(12)=10000(0.87449)^{12}=2000 \text{approximately}

b) P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}

c) P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}

d) P(12)=10000(0.87449)^{12+12}=400 \text{approximately}

So a is the function we want.

Also another way to look at this:

P(t)=10000(.2)^t where t is in years.

P(t)=10000(.2^\frac{1}{12})^t where t is in months.

And .2^\frac{1}{12}=0.874485 \text{approximately}

8 0
3 years ago
How does the value of log2^100 compare with the value of log6^20?
ipn [44]
It is A.. Value of log2^100 is about 4 times the value of log6^20
6 0
3 years ago
Read 2 more answers
Help me i don’t know how to do this
barxatty [35]

Hi!

y = 16

x = 32

<h3 /><h3>In 30-60-90 triangles, the hypotenuse is double the length of the shortest leg, and the longer leg is \sqrt{3} times the shorter leg.</h3>

We are given the longest leg. To find the shortest leg, we must divide the longest leg value by \sqrt{3}

\cfrac{16\sqrt{3} }{\sqrt{3} }

The radicals cancel out and we are left with 16.

The length of y (the shortest leg) is 16.

Now, we also know that the hypotenuse is double the shortest leg. The shortest leg is 16, so if we double that, it's 32.

8 0
2 years ago
C = 200 – 7x + 0.345x2
fenix001 [56]

The way to find the domain and the range is illustrated below.

<h3>How to illustrate the information?</h3>

From the information, c(x) = 200 - 7x + 0.345x². It should be noted that the domain is the set of x-values that are feasible.

The range is the set of possible results for c(x). They are the possible costs. You can derive this from the fact that c(x) is a parabola and you can draw it, for which you can find the vertex of the parabola, the roots, the y-intercept, the shape.

You can substitute some values for x to help you, for example:

x      y

0    200

1    200 -7 +0.345 = 193.345

2    200 - 14 + .345 (4) = 187.38

3    200 - 21 + .345(9) = 182.105

4    200 - 28 + .345(16) = 177.52

5    200 - 35 + 0.345(25) = 173.625

6    200 - 42 + 0.345(36) = 170.42

10  200 - 70 + 0.345(100) =164.5

11 200 - 77 + 0.345(121) = 164.745

If the functions do not have real roots, then the costs never decrease to 0. The function starts at c(x) = 200, decreases until the vertex, (x =10, c=164.5) and starts to increase.

Then the range goes from 164.5 to infinity, limited to the solution for x = positive integers.

Learn more about domain on:

brainly.com/question/2264373

#SPJ1

Complete question:

the daily production cost, c, for x units is modeled by the equation: c = 200 – 7x 0.345x² explains how to find the domain and range of c.

4 0
1 year ago
PLEASE HELPPPPPP EASY
lianna [129]
The answer to your question is D
5 0
2 years ago
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