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Arlecino [84]
2 years ago
6

4(x − 3) − 5(x + 1) = 3 Which of the following algebraic properties is not needed to solve this equation?

Mathematics
1 answer:
skelet666 [1.2K]2 years ago
7 0
Hi there!

To solve this problem, you need to use the distributive property. You distribute the 4 inside the parenthesis as well as the 5.

Hope this helps!
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Car rentals involve a $130 flat fee and an additional cost of $31.67 a day. what is the maximum number of days you can rent a ca
Vladimir79 [104]
Represent the number of days by x. With this representation, the variable cost of the rental is 31.67x. The total cost is the sum of the fixed and variable costs. This value should not be more than $500. The equation below shows the relationship.
                                          130 + 31.67x ≤ 500

Solving for x gives x ≤ 11.68
Thus, the maximum number of days to rent the car is only 11 days. 
6 0
2 years ago
Read 2 more answers
Solve the equations to find the number and type of solutions
luda_lava [24]

Answer:

This has one real solution, x=4

Step-by-step explanation:

8 - 4x = 0

Add 4x to each side

8 - 4x+4x = 0+4x

8 =4x

Divide each side by 4

8/4 = 4x/4

2 =x

This has one real solution, x=4

6 0
3 years ago
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What is the equation of the line that is parallel to the line 5x + 2y = 12 and passes through the point (−2, 4)? y = – x – 1y =
Firlakuza [10]
The equation of the line parallel to 5x+2y=12 and passes through the point (-2, 4) is equal to y= -5/2x - 1
7 0
2 years ago
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A computer valued at $6500 depreciates at the rate of 14.3% per year. Find the value of the computer after 3 years
Norma-Jean [14]
After 3 years the computer will be valued at 3711.5. I'm sure you can round it up to 3712.

Hope this helps :)
6 0
3 years ago
SOLVING EACH SYSTEM BY ELIMINATION <br><br><br> x+3y=5 x=-6y+14<br><br><br> HELP
nata0808 [166]

Answer:

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

x=-4,\:y=3

Step-by-step explanation:

Given the system of the equations

x+3y=5;\:x=-6y+14

solving by elimination method

\begin{bmatrix}x+3y=5\\ x=-6y+14\end{bmatrix}

\mathrm{Arrange\:equation\:variables\:for\:elimination}

\begin{bmatrix}x+3y=5\\ x+6y=14\end{bmatrix}

x+6y=14

-

\underline{x+3y=5}

3y=9

\begin{bmatrix}x+3y=5\\ 3y=9\end{bmatrix}

solve 3y=9 for y:

3y=9

\frac{3y}{3}=\frac{9}{3}

y=3

\mathrm{For\:}x+3y=5\mathrm{\:plug\:in\:}y=3

Solve x+3\cdot \:3=5 for x:

x+3\cdot \:3=5

x+9=5

x=-4

Therefore,

\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}

x=-4,\:y=3

6 0
2 years ago
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