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nalin [4]
3 years ago
13

Distance to the earth's center kilometer

Mathematics
1 answer:
Veseljchak [2.6K]3 years ago
6 0

Answer:

its 6,371 in kilometers

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PLEASE HELP! WILL MARK BRAINLIEST
Veseljchak [2.6K]

The value of x is 3, while the length of the rectangle is 30 units and the width is 24 units.

<h3>How to determine the dimensions?</h3>

The given parameters are:

Base = 5(x+3)

Height = 2(x+9)

Perimeter = 108

The perimeter of a rectangle is

P = 2 *(Base + Height)

So, we have:

2 *(5(x + 3) + 2(x + 9)) = 108

Divide both sides by 2

5(x + 3) + 2(x + 9) = 54

Open the brackets

5x + 15 + 2x + 18 = 54

Evaluate the like terms

7x = 21

Divide by 7

x = 3

Substitute x = 3 in Base = 5(x+3) and Height = 2(x+9)

Base = 5(3+3) = 30

Height = 2(3+9) = 24

Hence, the value of x is 3, while the length of the rectangle is 30 units and the width is 24 units.

Read more about perimeter at:

brainly.com/question/24571594

#SPJ1

5 0
1 year ago
Your favorite restaurant offers a total of 14 desserts, of which 8 have ice cream as a main ingredient and 12 have fruit as a ma
djyliett [7]
Your favorite restaurant offers a total of 13 desserts, of which 11 have ice cream as a main ingredient and 9 have fruit as a main ingredient. Assuming that all of them have either...
3 0
3 years ago
What is the area of this parallelogram? PLEASE HELP
zimovet [89]

Answer:

B

Step-by-step explanation:

the area of a parallelogram is the length of one of the sides multiplied with the length of the height on this side.

so, in this case it is

4×5 = 20 cm²

5 0
2 years ago
Fritz is in charge of buying uniform for his baseball team. one uniform costs $20. there are 14 players on his team. how much wi
tekilochka [14]
If 1 uniform costs $ 20....and there are 14 players....then Fritz would spend 
(20 x 14) = $ 280.....thats if each player only gets 1 uniform .

8 0
3 years ago
Question :-
DochEvi [55]

\qquad\qquad\huge\underline{{\sf Answer}}

Let's solve ~

\qquad \sf  \dashrightarrow \:  \bigg( \dfrac{1}{ \sqrt{7 +  \sqrt{5} } }  +  \sqrt{7 +  \sqrt{5} } \bigg) {}^{2}

\qquad \sf  \dashrightarrow \:  \bigg( \dfrac{1}{ \sqrt{7 +  \sqrt{5} } } \bigg ) {}^{2}  +  \bigg( \sqrt{7 +  \sqrt{5} } \bigg) {}^{2}  + 2 \cdot \dfrac{1}{ \sqrt{7 +  \sqrt{5} } }  \cdot \sqrt{7 +  \sqrt{5} }

\qquad \sf  \dashrightarrow \: \dfrac{1}{7 +  \sqrt{5} }  + 7 +  \sqrt{5}  + 2

\qquad \sf  \dashrightarrow \: \dfrac{49 + 7 \sqrt{5} + 7 \sqrt{5} + 5 + 14 + 2 \sqrt{5}   }{7 +  \sqrt{5} }

\qquad \sf  \dashrightarrow \: \dfrac{68+ 16 \sqrt{5}    }{7 +  \sqrt{5} }

5 0
1 year ago
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