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Natali [406]
2 years ago
5

The number of water bottles used during a teams football practice varies directly with the temperature. If a team uses 75 water

bottles when the temperature is 60 degrees, what is the temperature if they use 120 bottles?

Mathematics
1 answer:
AysviL [449]2 years ago
6 0
To find the cost of 1 water bottle you divide 60 degrees into 75 (75/60) which gives you 1.25 degrees for each water bottle. To find 120 water bottles temperature multiply the 1.25 degrees for 1 water bottle times 120 (120x1.25). 120x1.25=150 The answer would be 150degrees
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2 years ago
A department store sells small boxes in which customers can store potpourri. The diagram below shows one of the boxes. The diagr
larisa86 [58]
The lateral area is expressed as the product of the perimeter of the base and the height. While the surface area is the sum of all areas. The lateral area and the surface area of the box is calculated as follows:

LA = Ph 
P = 2l + 2w = (2 x 17) + (2 x 5) = 44 cm
LA = 44 cm x 3.12 cm = 137.28 cm

SA = 2(lw + wh + lh)
SA = 2[(17x5) + (5x3.12) + (17x3.12)]
SA = 307.28 cm

Thus, the lateral area and the surface area are 137 cm and 307 cm, respectively.
8 0
3 years ago
If f (x) = x + 7 and g (x) = StartFraction 1 Over x minus 13 EndFraction, what is the domain of (f circle g) (x)?
Rashid [163]

Answer:

R-{13}

Step-by-step explanation:

We are given that

f(x)=x+7

g(x)=\frac{1}{x-13}

We have to find the domain of fog(x).

fog(x)=f(g(x))

fog(x)=f(\frac{1}{x-13})

fog(x)=\frac{1}{x-13}+7=\frac{1+7x-91}{x-13}=\frac{7x-90}{x-13}

Domain of f(x)=R

Because it is linear function.

Domain of g(x)=R-{13}

Because the g(x) is not defined at x=13

fog(x) is not defined at x=13

Therefore, domain of fog(x)=R-{13}

6 0
3 years ago
Read 2 more answers
A basketball team has won 0.625 of its games. If the team has played 24 games, how many games has the team won?
BlackZzzverrR [31]

Answer:

It has won 15 games.

Step-by-step explanation:

24*.625=15

6 0
10 months ago
Find the vector projection of B onto A if A = 5i + 11j – 2k,B = 4i + 7k​
valkas [14]

Answer:

\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\

Step-by-step explanation:

Given A = 5i + 11j – 2k and B = 4i + 7k​, the vector projection of B unto a is expressed as proj_ab = \dfrac{b.a}{||a||^2} * a

b.a = (5i + 11j – 2k)*( 4i + 0j + 7k)

note that i.i = j.j = k.k  =1

b.a = 5(4)+11(0)-2(7)

b.a = 20-14

b.a = 6

||a|| = √5²+11²+(-2)²

||a|| = √25+121+4

||a|| = √130

square both sides

||a||² = (√130)

||a||²  = 130

proj_ab = \dfrac{6}{130} * (5i+11j-2k)\\\\proj_ab = \frac{30}{130} i+\frac{11}{130} j-\frac{12}{130} k\\\\proj_ab = \frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\\\

<em>Hence the projection of b unto a is expressed as </em>\frac{3}{13} i+\frac{11}{130} j-\frac{6}{65} k\\<em></em>

7 0
3 years ago
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