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grandymaker [24]
3 years ago
12

Distance between (1,-2) and (4,4)

Mathematics
1 answer:
Elenna [48]3 years ago
8 0
\bf\textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ 1}}\quad ,&{{ -2}})\quad 
%  (c,d)
&({{ 4}}\quad ,&{{ 4}})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
d=\sqrt{[4-1]^2+[4-(-2)]^2}\implies d=\sqrt{(4-1)^2+(4+2)^2}
\\\\\\
d=\sqrt{3^2+6^2}\implies d=\sqrt{45}\implies d=\sqrt{3\cdot 3\cdot 5}\implies d=\sqrt{3^2\cdot 5}
\\\\\\
d=3\sqrt{5}
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Martas school has 325 desks for 13 classrooms . If the desks are shared evenly how many desks will each classroom have
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Step-by-step explanation:

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3 years ago
A movie theater has a seating capacity of 329. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults.
kipiarov [429]

Answer:

170 children

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85 adults

Step-by-step explanation:

Given

Let:

C = Children; S = Students; A = Adults

For the capacity, we have:

C + S + A = 329

For the tickets sold, we have:

5C + 7S + 12A = 2388

Half as many as adults are children implies that:

A = \frac{C}{2}

Required

Solve for A, C and S

The equations to solve are:

C + S + A = 329 -- (1)

5C + 7S + 12A = 2388 -- (2)

A = \frac{C}{2} -- (3)

Make C the subject in (3)

C = 2A

Substitute C = 2A in (1) and (2)

C + S + A = 329 -- (1)

2A + S + A = 329

3A + S = 329

Make S the subject

S = 329 - 3A

5C + 7S + 12A = 2388 -- (2)

5*2A + 7S + 12A = 2388

10A + 7S + 12A = 2388

7S + 22A = 2388

Substitute S = 329 - 3A

7(329 - 3A) + 22A = 2388

2303 - 21A + 22A = 2388

2303 +A = 2388

Solve for A

A = 2388 - 2303

A = 85

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C = 2 * 85

C = 170

Recall that: S = 329 - 3A

S = 329 - 3 * 85

S = 329 - 255

S = 74

Hence, the result is:

C = 170

S = 74

A = 85

8 0
3 years ago
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