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just olya [345]
3 years ago
5

Describe real life situations that use parallel lines ?????

Mathematics
2 answers:
Anna71 [15]3 years ago
5 0
A bed making a frame
A tv
A window
These are my answers
azamat3 years ago
5 0
Roads.
construction workers are ordered to build a new road, they need to use parallel lines so the roads aren't confusing and so there aren't any car accidents. that could be a response
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The length of a rectangle is 5 feet and the width is 8 feet.what is the area of the rectangle​
aliya0001 [1]

Answer:

40 ft squared

Step-by-step explanation:

To calculate area of a rectangle, you use the formula length times width, or base times height.

5 x 8 = 40 feet squared

5 0
3 years ago
Systems of elimination for<br> 2w-3q=8<br> 3w-7q=7
zlopas [31]

i guess is by multiplying 3by the first equation and multiply 2by the second equation

8 0
2 years ago
A water truck is filling a swimming pool. The equation that represents this relationship is y = 19.75x where y is the number of
Degger [83]

Answer:

Step-by-step explanation:

1. true

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5 0
2 years ago
A car's position is given by s(t) = {3 – 5t? + 7t hundreds of meters with t in minutes.
Nadya [2.5K]

Answer:

A) (1 s, 2.3 s)

B) (-4 m/s², 3.8 m/s²)

Step-by-step explanation:

The car's position which is the distance is given by the equation;

s(t) = t³ - 5t² + 7t

A) Velocity is the first derivative of the distance. Thus;

v(t) = ds/dt = 3t² - 10t + 7

At v = 0, we have;

3t² - 10t + 7 = 0

Using quadratic formula, we have;

t = 1 and t = 2.3

Thus, time at velocity of 0 is t = (1 s, 2.3 s)

B) acceleration is the derivative of the velocity. Thus;

a(t) = dV/dt = 6t - 10

At velocity of 0, we got t = 1 and t = 2.3

Thus;

a(1) = 6(1) - 10 = -4 m/s²

a(2.3) = 6(2.3) - 10 = 3.8 m/s

Thus, a(t) at v = 0 gives; (-4 m/s², 3.8 m/s²)

8 0
2 years ago
Y is directly proportional to square root of x<br> If y=56 when x=49 find,<br> y when x=81
STALIN [3.7K]

\qquad \qquad \textit{direct proportional variation} \\\\ \textit{\underline{y} varies directly with \underline{x}}\qquad \qquad \stackrel{\textit{constant of variation}}{y=\stackrel{\downarrow }{k}x~\hfill } \\\\ \textit{\underline{x} varies directly with }\underline{z^5}\qquad \qquad \stackrel{\textit{constant of variation}}{x=\stackrel{\downarrow }{k}z^5~\hfill } \\\\[-0.35em] \rule{34em}{0.25pt}

\stackrel{\begin{array}{llll} \textit{"y" directly}\\ \textit{proportional to }\sqrt{x} \end{array}}{y = k\sqrt{x}}\qquad \textit{we know that} \begin{cases} y = 56\\ x = 49 \end{cases}\implies 56=k\sqrt{49} \\\\\\ 56=7k\implies \cfrac{56}{7}=k\implies 8=k~\hfill \boxed{y=8\sqrt{x}} \\\\\\ \textit{when x = 81, what is "y"?}\hfill y=8\sqrt{81}\implies y=8(9)\implies y=72

7 0
2 years ago
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