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Ad libitum [116K]
2 years ago
15

The perimeter of the rectangle below is 84 units. Find the length of side VW.

Mathematics
1 answer:
Anestetic [448]2 years ago
8 0

Answer:

18.

Step-by-step explanation:

1) formula of the perimeter given in the condition is:

p=2(VW+XV) or p=2(VY+XY).

Note, VW=XY and XV=VY;

2) if to substitute p=84, VW=2z+2 and XV=3z into the formula above, then

2(2z+2+3z)=84, ⇔ 5z+2=42; ⇒ z=8.

3) if z=8, then VW=2z+2=2*8+2=18.

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Employee x earned $5250 in 3 weeks. at this rate how much will employee x earn in 8 weeks
iVinArrow [24]

Answer:

$14000

Step-by-step explanation:

Given parameters:

Amount earned by employee x  = $5250

Time duration  = 3weeks

Unknown:

Amount earned in 8 weeks = ?

Solution:

Let us first find the rate of employee x earning;

       Rate  = \frac{Amount earned}{Time duration}  

Insert parameters;

      Rate  = \frac{5250}{3}  = $1750/week

So,

  In 8 weeks;

      Amount earned  = rate x time duration

      Amount earned  =  $1750/week x 8weeks

      Amount earned  = $14000

8 0
2 years ago
Someone please help me with two answer choices
otez555 [7]

Answer:

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7 0
3 years ago
Which value of a would make the following expression completely factored?
Margarita [4]

Answer:

when it ur first question and u make it hard

Step-by-step explanation:

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6 0
3 years ago
Greatest 3 digit multiple of 33 using 3 diffrent digits
mel-nik [20]

Answer:

957

The answer is this because the next 3 digit multiple of 33 would be 990.

4 0
3 years ago
Read 2 more answers
A trailer will be used to transport several 40-pound crates to a store. The greatest amount of weight that can be loaded into th
docker41 [41]

Given :

A trailer will be used to transport several 40-pound crates to a store.

The greatest amount of weight that can be loaded into the trailer is 1,050 pounds.

An 82-pound crate has already been loaded onto the trailer.

To Find :

The greatest number of 40-pound crates that can be loaded onto the trailer.

Solution :

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Let, number of 40 pounds crates that can be loaded are x.

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Since, crate cannot be in fraction, so maximum crate that can be loaded is 24.

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7 0
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