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2 years ago

A pool covers costs about $17.50 per square meter. Calculate the cost of covering the Children's pool.

2 answers:

8 0

First calculate the area of the pool, which is 15 x 15, = 225 sq meter,
and just multiply the cost for each sq meter by the area,
17.5 x 225
therefore the cost will be $3937.5

arsen [322]2 years ago

6 0

In order to answer your question, you must find the area of the children's pool. Since it is mentioned that the pool's shape is a square and has a side of 15m, you have to find the area of the square, which is s^2. 15m^2 is equal to 225 square meters. Since one square meter costs $17.50, the total cost of covering the pool is 17.50*225, which equals to $3937.50.

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Write the equation of the circle whose center is (-4,8) and has a diameter of 18

svetoff [14.1K]

Answer:

(x + 4)^2 + (y - 8)^2 = 81

(x + 4)^2 + (y - 8)^2 = 9^2 depending on how your teacher wants it written.

Step-by-step explanation:

The standard form for a circle is

(x + h)^2 + (y + k)^2 = r^2

r is the radius.

You are given the diameter

r = d/2

r = 18/2

r = 9

So you already have the right hand side of the equation

(x + h)^2 + (y + k)^2 = 9*2

(x + h)^2 + (y + k)^2 = 81

You basically have h and k as well. They come from the center point.

h = 4

k = - 8

So the equation of the circle (and the answer) is

(x + 4)^2 + (y - 8)^2 = 81

One question remains. Why do the x and y values change signs? If you know what the distance formula is, then what you are finding is the distance r of all points on the circle from the center of the circle.

It is the distance formula that is actually the formula for the circle.

6 0

2 years ago

A+9 write this in a verbal expression

balandron [24]

Nine more than a variable "a".

7 0

2 years ago

Let T be the plane-2x-2y+z =-13. Find the shortest distance d from the point Po=(-5,-5,-3) to T, and the point Q in T that is cl

GaryK [48]

Answer:

d=10u

Q(5/3,5/3,-19/3)

Step-by-step explanation:

The shortest distance between the plane and Po is also the distance between Po and Q. To find that distance and the point Q you need the perpendicular line x to the plane that intersects Po, this line will have the direction of the normal of the plane $n=(-2,-2,1)$ , then r will have the next parametric equations:

$x=-5-2\lambda\\y=-5-2\lambda\\z=-3+\lambda$

To find Q, the intersection between r and the plane T, substitute the parametric equations of r in T

$-2x-2y+z =-13\\-2(-5-2\lambda)-2(-5-2\lambda)+(-3+\lambda) =-13\\10+4\lambda+10+4\lambda-3+\lambda=-13\\9\lambda+17=-13\\9\lambda=-13-17\\\lambda=-30/9=-10/3$

Substitute the value of $\lambda$ in the parametric equations:

$x=-5-2(-10/3)=-5+20/3=5/3\\y=-5-2(-10/3)=5/3\\z=-3+(-10/3)=-19/3\\$

Those values are the coordinates of Q

Q(5/3,5/3,-19/3)

The distance from Po to the plane

$d=\left| {\to} \atop {PoQ}} \right|=\sqrt{(\frac{5}{3}-(-5))^2+(\frac{5}{3}-(-5))^2+(\frac{-19}{3}-(-3))^2} \\d=\sqrt{(\frac{5}{3}+5))^2+(\frac{5}{3}+5)^2+(\frac{-19}{3}+3)^2} \\d=\sqrt{(\frac{20}{3})^2+(\frac{20}{3})^2+(\frac{-10}{3})^2}\\d=\sqrt{\frac{400}{9}+\frac{400}{9}+\frac{100}{9}}\\d=\sqrt{\frac{900}{9}}=\sqrt{100}\\d=10u$

7 0

2 years ago

Sue Mitchell weighed 8 lb. 13 oz. when she was born. She now weighs 93 lb. 5

ra1l [238]

Answer:

Weight gain = 84 lb 5 oz

Average weight gain = 6 lb 5 oz

Step-by-step explanation:

lb means pounds

16 oz = 1 pound

Convert the weight to pounds

Weight when she was born = 8 lb. 13 oz

= 8 pounds + 0.8125 pounds

= 8.8125 pounds

Present weight = 93 lb. 5

= 93 pounds + 0.3125 pounds

= 93.3125 pounds

How much has she gained since birth

Weight gain = Present weight - Weight when she was born

= 93.3125 pounds - 8.8125 pounds

= 84.5 pounds

= 84 lb 5 oz

Weight gain = 84 lb 5 oz

what has been her average weight for each year if she is 13 year old

Average weight gain = Weight gain / 13

= 84.5 pounds / 13

= 6.5 pounds

= 6 lb 5 oz

Average weight gain = 6 lb 5 oz

8 0

2 years ago

In 1980, there were 67 new productions on Broadway. In 1987, there were 31 new productions on Broadway. Find the average rate of

AlladinOne [14]

In 1980, there were 67 new productions on Broadway. In 1987, there were 31 new productions on Broadway. Find the average rate of change in the number of new productions.

6 0

3 years ago

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