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Mars2501 [29]
4 years ago
13

Which best describes the relationship between the line that passes through (7, 1) and (10, 5) and the line that

Mathematics
1 answer:
Lynna [10]4 years ago
6 0
4/3 ig I kept I did y2-y1 division
x2-x1 division
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Consider the equation
timama [110]
27 and 64 lmk if it’s wrong
6 0
3 years ago
At a movie theater they sell hotdogs and popcorn. The first hour they sold 15 popcorns and 14 hotdogs for a total of $84. The se
Delicious77 [7]

Answer:

1 hotdog cost $2.25

1 popcorn cost $3.5

Step-by-step explanation:

Let the price of hotdog be and that of popcorn be p

For the first hour;

15p + 14h = 84 •••• (i)

for the second hour;

18p + 16h = 99 •••••(ii)

so we want to solve these two equations simultaneously

In equation 1,

14h = 84-15p

h = (84-15p)/14

Simply insert this into equation ii

18p + 16(84-15p)/14 = 99

Multiply through by 14

252p + 1344 - 240p = 1386

1386-1344 = 252p-240p

12p = 42

p = 42/12

p = $3.5

to get cost of hotdogs;

h = (84-15p)/14

Substitute p = 3.5

h = (84-52.5)/14

h = 31.5/14

h = 2.25

8 0
3 years ago
What is the value of x in the equation 2x + 3y = 36, when y = 6? <br>08 <br>9 <br>27 <br>36​
lara31 [8.8K]
2x+3(6)=36
2x+18=36
subtract 18 both sides
2x=18
x=9
4 0
3 years ago
Read 2 more answers
PLZZZZ HELPPPPPPP!!!!
Vladimir79 [104]

Answer: The height of the building is 50.75 feet.

Step-by-step explanation:

The ratio between the height of the object and the casted shadow must be equal for all the objects, as the angle at which the source if light impacts them is the same.

For the person, we know that it is 5.8ft tall, and the shadow is 3.2ft long.

The ratio will be: 5.8ft/3.2ft = 1.8125

Now, if H is the height of the building, and the shadow that the building casts is 28ft, we must have:

H/28ft = 1.8125

Now we can solve this for H.

H = 1.8125*28ft = 50.75 ft

Then the height of the building is 50.75 feet.

3 0
3 years ago
Algebra II help I'm struggling
Dmitry [639]
Answer is B

total cost = $30 + 0.5 *(miles driven)
.................
7 0
4 years ago
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