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2 years ago

PLS PLS PLS HELP I HAVE 2 TESTS DUE BY MIDNIGHT LOLZ

Mathematics

2 answers:

sveticcg [70]2 years ago

8 0

The initial value is 3)

cricket20 [7]2 years ago

4 0

Answer:

Step-by-step explanation:

The graph is moving up by 3 units and right by 7 units. The 1st point and 2nd point are 7 units apart, which is where I got my answer.

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In 2003 a town’s population was 1431. By 2007 the population had grown to 2134. Assume the population is changing linearly.How m

sergiy2304 [10]

We will have the following:

First, we will have that the population change by:

$\Delta P=2134-1431\Rightarrow\Delta P=703$

So, the population changed by 703 people.

It took the follwoing number of years to change from 1431 to 2134:

$\Delta y=2007-2003\Rightarrow\Delta y=4$

So, it took 4 years to changes from 1431 to 2134.

$m=\frac{2134-1431}{2007-2003}\Rightarrow m=175.75$

The average population growth per year is of 175.75 people per year.

The population in 2000 would have been:

$P_{2000}=\frac{2000\cdot1431}{2003}\Rightarrow P_{2000}=1428.856715\ldots$

$\Rightarrow P_{2000}\approx1429$

So, the population in 2000 would have been approximately 1429 people.

We will have that the equation for the population would be:

$P=(\frac{2134-1431}{2007-2003})t+1429\Rightarrow P=\frac{703}{4}t+1429$

So, the equation would be:

$P=175.75t+1429$

Now, we predict the number of people of the town in 2014:

$P_{2014}=175.75(14)+1429\Rightarrow P_{2014}=3889.5$

$\Rightarrow P_{2014}\approx3890$

So, there will be approximately 3890 people of the town in 2014.

3 0

1 year ago

Describe lengths of three segments that could not be used to form a triangle. Segments with lengths of 5 in., 5 in., and ______

Vesna [10]

Answer:

$c = 11$

Step-by-step explanation:

Given

Let the three sides be a, b and c

Such that:

$a = b= 5$

Required

Find c such that a, b and c do not form a triangle

To do this, we make use of the following triangle inequality theorem

$a + b > c$

$a + c > b$

$b + c > a$

To get a valid triangle, the above inequalities must be true.

To get an invalid triangle, at least one must not be true.

Substitute: $a = b= 5$

$5 + 5 > c$ $===>$ $10 > c$

$5 + c > 5$ $===>$ $c > 5 - 5$ $===>$ $c > 0$

The results of the inequality is: $10 > c$ and $c > 0$

Rewrite as: $c > 0$ and $c < 10$

$0 < c < 10$

This means that, the values of c that make a valid triangle are 1 to 9 (inclusive)

Any value outside this range, cannot form a triangle

So, we can say:

$c = 11$ , since no options are given

8 0

2 years ago

Which system is equivalent to the system shown and could be used to solve the system of equations correctly using the eliminatio

Bumek [7]

Answer:

The solution is (2, -2)

Step-by-step explanation:

In order to solve this, multiply the second equation by -5 and then add through.

5x - 4y = 18

-5x - 15y = 20

------------------

-19y = 38

y = -2

Now that we have the value of y, we can solve for x.

x + 3y = -4

x + 3(-2) = -4

x - 6 = -4

x = 2

5 0

3 years ago

1/2 = 2m what does m equal

max2010maxim [7]

1/4 time 1/4 equals a half...or you can think about it as 25 plus 25 equals 50, (1/2 of a dollar)

8 0

3 years ago

One number was divided by 2, followed by this quotient being multiplied by 4. next, the product was added to 9. the total sum wa

grin007 [14]

Answer:

Step-by-step explanation:

First, turn the word problem into an equation:

n(a number) was divided by 2

$\frac{n}{2}$

followed by this quotient beying multiplied by 4

$\frac{n}{2}$ *4

the product was added to 9

$\frac{n}{2}$ *4+9

the total sum was 25

$\frac{n}{2}$ *4+9 = 25

use PEMDAS to solve the equation

subtract 9 on both sides

$\frac{n}{2}$ *4 = 16

cancel the 2 and the 4 using factorization

n*2 = 16

divide 2 on both sides

n = 8

7 0

2 years ago

PLS PLS PLS HELP I HAVE 2 TESTS DUE BY MIDNIGHT LOLZ ​

PLS PLS PLS HELP I HAVE 2 TESTS DUE BY MIDNIGHT LOLZ