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STALIN [3.7K]
3 years ago
5

Yossi used to have a square garage with 334ft^2 of floor space. He recently built an addition to it. The garage is still square,

but now it has 50% more floor space. What was the length of one side of the garage originally? What is the length of one side of the garage now? What was the precent increase in the length on one side?
​
Mathematics
1 answer:
vaieri [72.5K]3 years ago
4 0

Answer:

Increase in the length of one side = 3.8 ft

Step-by-step explanation:

Yossi has garage whose area is 280 ft²

After rebuilt their is increment in area by 50%

New Area = 280 × 150% = 420 ft²

Also we know that Area of Square = (Length) × (Length)

⇒ Length of Newly built garage = √420 = 20.5 ft

and Length of Originally garage = √280 = 16.7 ft

⇒ Increase in the length of one side = 20.5 - 16.7 = 3.8 ft

-

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Step-by-step explanation:

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3x+3x2+2x2+2x

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3 years ago
Evaluate. <br> 6[5(3-9)-1]+2/7(8-2)+4
GREYUIT [131]
If you would like to evaluate 6 * [5 * (3 - 9) - 1] + 2 / 7 * (8 - 2) + 4, you can calculate this using the following steps:

6 * [5 * (3 - 9) - 1] + 2 / 7 * (8 - 2) + 4 = 6 * [5 * (-6) - 1] + 2/7 * 6 +  4 = 6 * [- 30 - 1] + 12/7 + 4 = 6 * [- 31] + 12/7 + 4 = - 186 + 12/7 + 4 = - 182 + 12/7 = - 1274/7 + 12/7 = - 1262/7

The correct result would be - 1262/7.
6 0
4 years ago
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A water tanks can hold 234 cubic feet of water.How many gallons can the water tank hold
Veseljchak [2.6K]

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Hence, the water tank can hold 1750.32 gallons of water.

4 0
3 years ago
Please help and explain! *Serious answers only* 30 points :)
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4 0
3 years ago
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The center of a hyperbola is located at the origin. One focus is located at (−50, 0) and its associated directrix is represented
leva [86]

The equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

The center of a hyperbola is located at the origin that means at (0, 0) and one of the focus is at (-50, 0)

As both center and the focus are lying on the x-axis, so the hyperbola is a horizontal hyperbola and the standard equation of horizontal hyperbola when center is at origin: \frac{x^{2}}{a^{2}}  - \frac{y^{2}}{b^{2}}    = 1

The distance from center to focus is 'c' and here focus is at (-50,0)

So, c= 50

Now if the distance from center to the directrix line is 'd', then

d= \frac{a^{2}}{c}

Here the directrix line is given as : x= 2304/50

Thus, \frac{a^{2}}{c}  = \frac{2304}{50}

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⇒ b² = 50² - 48² (By plugging c=50 and a = 48)

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⇒ b = √196 = 14

So, the equation of the hyperbola is : \frac{x^{2}}{48^2}  - \frac{y^{2}}{14^2}  = 1

5 0
3 years ago
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