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3 years ago

the perimeter of a pool table is 30 ft.The table is twice as long as it is wide.what is the length of the pool table.

1 answer:

4 0

Perimeter is legnth +legnth + width + width or
P=2L+2W

L=2W
subsitute L=2W for L in P=2L+2W

P=2(2W)+2W
P=4W+2W
P=6W

we know that Perimiter=30 so
30=6W
divide both sides by 6
5=Width

subsitute W=5 for W in L=2W
L=2(5)
L=10

to check
10=2(10)+2(5)
30=20+10
30=30
checks

Legnght=10
Width=5

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Wittaler [7]

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3 years ago

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JulijaS [17]

Subtract 3x from both sides:

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3 years ago

Which of the lines is perpendicular to the line shown in the graph

Nitella [24]

Answer:

The line passing through (-8, 10) and (-1, 4).

Step-by-step explanation:

Two lines are perpendicular if the product of their slopes is -1. The slope of the line in the picture is $7\over 6$ , so we should find a line with slope of $-6\over 7$ .

Note that the slope of the line in the last option is $-7\over 6$ .

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2 years ago

Suppose you are determining the growth rate of two species of plants. Species A ls 25 cm tall and grows 3 cm per month. Species

exis [7]

Answer: Last Option

$H (m) = 25 + 3m\\H (m) = 10 + 8m$

Step-by-step explanation:

The initial height of the plant of species A is 25 cm and grows 3 centimeters per month.

If m represents the number of months elapsed then the equation for the height of the plant of species A is:

$H (m) = 25 + 3m$

For species B the initial height is 10 cm and it grows 8 cm each month

If m represents the number of months elapsed then the equation for the height of the plant of species B is:

$H (m) = 10 + 8m$

Finally, the system of equations is:

$H (m) = 25 + 3m\\H (m) = 10 + 8m$

The answer is the last option

8 0

3 years ago

Write out the first few terms of the series Summation from n equals 0 to infinity (StartFraction 2 Over 3 Superscript n EndFract

anyanavicka [17]

Answer:

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = 15/8$

Step-by-step explanation:

The sum you are trying to understand is this.

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n}$

Remember that in general when you have a geometric series

$\sum\limits_{n = 0}^{\infty} a*r^n$ you have that

$\sum\limits_{n = 0}^{\infty} a*r^n = \frac{a}{1-r}$ and that equality is true as long as $|r| < 1$ .

Therefore here we have

$\sum\limits_{n=0}^{\infty} \frac{2}{3^n}\frac{(-1)^n}{5^n} = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{3*5} \big)^n = \sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n$ and $\big|\frac{-1}{15} \big| = \frac{1}{15} < 1$

Therefore we can use the formula and

$\sum\limits_{n=0}^{\infty} 2* \big(\frac{-1}{15} \big)^n = \frac{2}{1-(-1/15)} = \frac{2}{1+1/15} = 30/16 = 15/8$

5 0

2 years ago