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Katarina [22]
3 years ago
15

HELP!!!

Mathematics
1 answer:
Eva8 [605]3 years ago
7 0

Answer:

False

Step-by-step explanation:

A ratio is not necessarily a comparison between quantities with different units, quantities with the same units can be in a ratio in mathematics.

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A 10ft by 20ft rectangular swimming pool is surrounded by a walkway of uniform width. If the total area of the walkway is 216ft^
Dmitry [639]

check the picture below.


so, we know the dimensions of the pool, is a 20x10, so its area is simply 200 ft², and we know the walkway is 216 ft², so the whole thing, including pool and walkway is really 200 + 216 ft².


now, as you see in the picture, the dimensions for the combined area is 20+2x and 10+2x, since the walkway is "x" long, therefore,


\bf \stackrel{length}{(20+2x)}\stackrel{width}{(10+2x)}=200+216\implies \stackrel{FOIL}{4x^2+60x+200}=200+216 \\\\\\ 4x^2+60x=216\implies \stackrel{dividing~by~4}{x^2+15x=54} \\\\\\ x^2+15x-54=0\implies (x+18)(x-3)=0\implies x= \begin{cases} -18\\ \boxed{3} \end{cases}


notice, it cannot be -18, since is a positive length unit.

5 0
3 years ago
Read 2 more answers
Please write it step by step: 7+6x-4=3x-9+7x
bazaltina [42]

Answer:

x = 3

Step-by-step explanation:

7 + 6x - 4 = 3x - 9 + 7x

6x -3x - 7x = -9 - 7 + 4

-4x = -12

-4x/-4 = -12/-4

x = 3

4 0
3 years ago
Read 2 more answers
Please please Help me!!!!!
pav-90 [236]

Answer:

19,23

Step-by-step explanation:

9+x-7≥ 21

Combine like terms

2+x≥21

Subtract 2 from each side

2+x-2 ≥21-2

x≥19

Any number greater than or equal to 19

3 0
3 years ago
Please help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Flura [38]

Answer:

#6 is 17 days

Step-by-step explanation:

This is how I answered it

1) 2750 = 550 + 125x

2) subtract 550 from each side

2750-550=2200

so the left over equation is:

2200 = 125x

3) divide each side by 125 to get x by itself

2200÷125= 17.6

4) you are left with: x = 17 days

6 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
2 years ago
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