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boyakko [2]
3 years ago
13

What is the answer for n + 4.5 - 0.3n -3

Mathematics
2 answers:
ratelena [41]3 years ago
8 0
1. Collect like terms

(n - 0.3n) + (4.5 - 3)

2. Simplify

0.7n + 1.5

Final Answer:

0.7n + 1.5
jekas [21]3 years ago
6 0

Answer: The answer is 0.7n + 1.5

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Answer:

960 sq units

Step-by-step explanation:

First: 22 times 16

Second: Multiply your first answer by 2

Third: 16 times 16

Fourth: Add your second and third answer for the solution

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3 years ago
A company earns P dollars by selling x items, according to the equation P(x)=-0.002x^2+5.5x-1000. How many items does the compan
san4es73 [151]
So.. check the picture below

that's when the profit gets maximized

\bf \qquad \textit{vertex of a parabola}\\ \quad \\

\begin{array}{lccclll}
P(x)=&-0.002x^2&+5.5x&-1000\\
&\uparrow &\uparrow &\uparrow \\
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6 0
3 years ago
Solve for x:<br> 5 + 3x = 8x - 15
Crazy boy [7]

Answer:

collect like terms

5+15=8x-3x

20=5x

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3 0
3 years ago
Read 2 more answers
Need help please help
azamat

Answer:

B

Hope you get it right! :)

Step-by-step explanation:

7 0
3 years ago
Wilma is 5 times as old as yvonne. in ten years, yvonne's age will be a third of wilma's age. how old are they now?
SVEN [57.7K]

Using a system of equations, it is found that Wilma is 50 years old and Yvonne is 10 years old.

<h3>What is a system of equations?</h3>

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the variables are given as follows:

  • Variable x: Wilma's age.
  • Variable y: Yvonne's age.

Wilma is 5 times as old as yvonne, hence:

x = 5y.

In ten years, yvonne's age will be a third of wilma's age, hence:

y + 10 = \frac{1}{3}(x + 10)

Since x = 5y:

y + 10 = \frac{1}{3}(x + 10)

y + 10 = \frac{1}{3}(5y + 10)

3y + 30 = 5y + 10

2y = 20.

y = 10.

Then:

x = 5y = 5 x 10 = 50.

Wilma is 50 years old and Yvonne is 10 years old.

More can be learned about a system of equations at brainly.com/question/24342899

#SPJ1

3 0
2 years ago
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