All categories

3 years ago

If 15 is added to a number the result is 56 less than twice the number

1 answer:

3 0

Let, the number = x
It would be: x + 15 = 2x - 56
2x - x = 15 + 56
x = 71

In short, Your Answer would be 71

Hope this helps!

You might be interested in

Select the postulate of equality or inequality that is illustrated 5 = 5

Oliga [24]

reflexive postulate of equality

7 0

3 years ago

3. The backyard of a home is a rectangle 25m by 30m. A garden of uniform width is to be

KiRa [710]

Answer:

Part A

The width of the garden is approximately 3.987 meters

Part B

Please find attached the drawing of the solution

Step-by-step explanation:

Part A

The given parameters are;

The dimensions of the backyard of the home = 25 m by 30 m

The width of the garden to be built in the yard = Uniform width

The shape of the grass left inside = Rectangular

The area of the grass at the center = The area of the garden

Let 'x' represent the width of the garden, we have;

The length of the rectangular grass area at the center, l = 30 - 2·x

The width of the rectangular area of grass at the center, w = 25 - 2·x

The area of the rectangular backyard, A = 30 m × 25 m = 750 m²

The area of the rectangular backyard, A = (The area of the garden) + (The area of the rectangle of grass inside)

The area of the rectangle of grass, GA = (30 - 2·x)·(25 - 2·x) = The area of the garden

The area of the rectangular backyard, A = 750 = (30 - 2·x)·(25 - 2·x) + (30 - 2·x)·(25 - 2·x) = 2 × (30 - 2·x)·(25 - 2·x) = 8·x² - 220·x + 1,500

∴ 750 = 8·x² - 220·x + 1,500

8·x² - 220·x + 1,500 - 750 = 0

8·x² - 220·x + 750 = 0

x = (220 ± √(220² - 4 × 8 × 750))/(2 × 8)

x ≈ 23.513, or x = 3.987

When x = 25.513, the width of the rectangle of grass inside, w = 25 - 2 × 23.513 = -22.026, which is not a natural (physically possible)

Therefore, the possible width of the garden, x ≈ 3.987

Part B

The drawing of the solution created with MS Visio is attached

6 0

2 years ago

Find the Surface Area and Volume of Sphere with radius 12 cm.

Gnom [1K]

$Radius:r=12cm\\\\Surface\ area:\\A=4\pi r^2\to A=4\pi\cdot12^2=4\pi\cdot144=576\pi\ (cm^2)\\\\Volume:\\V=\frac{4}{3}\pi r^3\to V=\frac{4}{3}\pi\cdot12^3=\frac{4}{3}\pi\cdot1728=4\pi\cdot576=2304\pi\ (cm^3)$

8 0

3 years ago

Read 2 more answers

-54 - 7r = -5(-2 - 5r)

makkiz [27]

-54-7r=10+25r
-7r-25r=10+54
-32r=64
r=64/-32
r=-2

3 0

3 years ago

15 POINTS!!! WILL GIVE BRAINLIEST PLEASE HELP ME!!!

Alina [70]

Answer:

#1Without the roof panels, the surface area (except for the open panel) is 176.25 sq. ft.

With the roof panels, the surface area (except for the open panel) is 196.67 sq. ft.

The area of the rectangular sections (the sides without gables) is given by 10*6 = 60 sq. ft. There are two of these sides, for a total of 2*60 = 120 sq. ft.

The gables are triangles, which have an area formula of

A = 1/2bh

For the gables, this is 1/2(1.5)(7.5) = 5.625 sq. ft. Since there are two of them, this makes 2*5.625 = 11.25 sq. ft.

The closed side under a gable has an area of 6*7.5 = 45 sq. ft.

The two roof panels have an area of (4 1/12)(2 1/2) each:

49/12(5/2) = 245/24

There are two of them, which makes 490/24 = 20.42 sq. ft.

Without the roof panels, we have 45+11.25+120 = 176.25 sq. ft.

With the roof panels, we have 45+11.25+120+20.42 = 196.67 sq. ft.

#2 Solar cells can transform solar energy into electric energy. In a sunny day, the average power the sun incides on earth is approximately 1,000 watts (heat). if we had a 100% efficient solar cell, it would mean that the total amount of sunlight energy would be transformed into electrical energy, i.e. 1,000 watts of power. Its efficiency could be 1000 watts(solar cells)/1000(sunlight)x100(to express it in percentage).

Since the solar cells that will be mounted in the roof will produce 41 watts per square meter, its efficiency is calculated by dividing 41 /1,000x100. This gives us an efficiency of 4.1%. In order to generate 1,000 watts of power, we will need to use a simple rule-of-three, described like the following:

if 41 watts are produced by the solar cells in 1 square meter, how many square meters will be needed to produce 1,000 watts?

i.e. 41 watts ----------- 1 sq meter

1,000 watts ------------ x

Finally, we can assume that the solar cell system will produce this amount of energy constantly during a full day cycle (using solar cells at day and batteries at night).

6 0

3 years ago