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2 years ago

Find the midpoint of the segment with the following endpoints. (10,6) and (6, 10)

Mathematics

1 answer:

lys-0071 [83]2 years ago

3 0

Midpoint = ( x1 + x2)/2, (y1+ y2)/2

Midpoint = (10 +6)/2, (6+10)/2

Midpoint = (8,8)

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PPLEASE HELP!!!!! WHAT IS THE INITIAL VALUE MEAN FOR THIS GRAPH/FUNCTION

Rudiy27

Amount saved:50I60I70l80l AND ON......
......weeks......:0..I1...l2..l3..l ..........AND ON
amount saved =x
weeks=y
after every y+10=x

6 0

3 years ago

Given a long algebraic equation, what are some strategies that you can use to make simplifying and evaluating the equation more

krok68 [10]

Answer:

See explanation

Step-by-step explanation:

Given a long algebraic equation, the like terms can be collected. When you collect like terms, you reduce the length of the algebraic equation.

After that, you can factorize the equation where possible. When you factorize the equation. It becomes quite easier to solve it efficiently.

6 0

3 years ago

Is X-13=x+1 true or false

Black_prince [1.1K]

Answer: FALSE.

Step-by-step explanation:

Given the following equation provided in the exercise:

$x-13=x+1$

You need to solve for the variable "x".

In order to solve for "x" you can folllow the steps shown below:

1. You must apply the Addition property of equality and add 13 to both sides of the equation. Then:

$x-13+(13)=x+1+(13)\\\\x=x+14$

2. Finally you need to apply the Subtraction property of equality and subtract "x" from both sides of the equation. So you get:

$x-(x)=x+14-(x)\\\\0=14\ (FALSE)$

Therefore, as you can observe, the given equation has no solutions.

3 0

3 years ago

Y= 2x + 6 for x = 0,1,2,3,4

algol [13]

x=0 y=6

x=1 y=8

x=2 y=10

x=3 y=12

x=4 y=14

8 0

3 years ago

The population of a colony of mosquitoes obeys the law of uninhibited growth. If there are 1000 mosquitoes initially and there a

Anettt [7]

Answer:

1) The size of the colony after 4 days is 6553 mosquitoes

2) After $t = 4.9\ days$

Step-by-step explanation:

To answer this question you must use the growth formula

$N = N_0e ^ {kt}$

Where

$N_0$ is the initial population of mosquitoes = 1000

t is the time in days

k is the growth rate

N is the population according to the number of days

We know that when t = 1 and $N_0=1000$ then N = 1600

Then we use these values to find k.

$1600 = 1000e ^{k(1)}\\\\\frac{1600}{1000} = e ^ k\\\\ln(\frac{1600}{1000}) = k\\\\k = 0.47$

Now that we know k we can find the size of the colony after 4 days.

$N = 1000e ^ {0.47(4)}\\\\N = 6553\ mosquitoes$

To know how long it should take for the population to reach 10,000 mosquitoes we must do N = 10000 and solve for t.

$10000 = 1000e ^ {0.47t}\\\\10 = e ^ {0.47t}\\\\ln(10) = 0.47t\\\\t = \frac{ln(10)}{0.47}\\\\t = 4.9\ days$

6 0

3 years ago