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vodomira [7]
3 years ago
14

Can someone please help someone who is good at geometry because I’m not sad face :(

Mathematics
1 answer:
Olenka [21]3 years ago
6 0
80 = 16x

80/16 = 5 :)
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The following graph represents the cost and revenue functions for custom-made computers. How many computers
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If 1/4 gallons of water are enough to fill 7/10 of the pool how much water is nessacary to fill in the entire pool
anyanavicka [17]
<h3>Answer:  5/14 gallons</h3>

============================================================

Work Shown:

1/4 = 0.25

7/10 = 0.70 = 70%

We are told that 0.25 gallons of water fills 70% of the pool

We want to find how many gallons of water (call it x) are needed to fill 100% of the pool.

Set up the proportion below and solve for x

(0.25 gallons)/(70% filled) = (x gallons)/(100% filled)

0.25/70 = x/100

0.25*100 = 70x ....... cross multiply

25 = 70x

70x = 25

x = 25/70

x = (5*5)/(5*14)

x = 5/14 gallons

----------------------------------------

An alternate method:

1/4 gallons = 7/10 pools filled

(10/7)*(1/4 gallons) = (10/7)*(7/10 pools filled)

(10*1)/(7*4) gallons = (10*7)/(7*10) pools filled

10/28 gallons = 1 pool filled

(5*2)/(14*2) gallons = 1 pool filled

5/14 gallons = 1 pool filled

In the second step, I multiplied both sides by 10/7 so that the "7/10" turns into "1".

----------------------------------------

Yet another alternate method:

1/4 = 10/40

7/10 = 28/40

We need 10/40 gallons of water to fill 28/40 of one pool. We can multiply both fractions by 40 to indicate that 10 gallons of water will fill up 28 pools.

So we have this ratio

10 gallons : 28 pools

Dividing both parts by 28 tells us how many gallons are needed for one pool

10 gallons : 28 pools

10/28 gallons : 28/28 pools

10/28 gallons : 1 pool

5/14 gallons : 1 pool

8 0
3 years ago
If y varies inversely as x and y= 2/5 when x= 2, find y when x= 1
Alik [6]

Answer:

y = 4/5

Step-by-step explanation:

Since, y varies inversely as x.

\therefore \: y  =  \frac{k}{x}  \\ (k = constant \: of \: proportionality) \\  \therefore \: xy = k...(1) \\ plug \: y =  \frac{2}{5}  \: and \: x = 2 \: in \: (1) \\  2\times \frac{2}{5}   = k \\  \implies \: k =  \frac{4}{5}  \\ substituting \: k=  \frac{4}{5}  \: in \: (1) \\ xy =  \frac{4}{5} ..(2) \\ this \: i s \: the \: equation \: of \: variation \\ plug \: x = 1 \: in \: (2) \\ 1 \times y =  \frac{4}{5}  \\  \huge \red{ \boxed{y =   \frac{4}{5}  }}

3 0
3 years ago
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