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telo118 [61]
3 years ago
7

Please answer this!! (In your own words.) The best answer gets brainliest! Find the area:

Mathematics
2 answers:
kipiarov [429]3 years ago
6 0

Answer:

97.5 sq. ft.

Step-by-step explanation:

See the given diagram with given measurements.

The area of the shaded portion is to be determined.

Now, we have to first find the area of the two right triangles formed at the left and right sides.

We know that the area of a right triangle = \frac{1}{2} \times (\textrm {Base})\times (\textrm {Height})

Then the area of the left side triangle = \frac{1}{2} \times 5 \times (12 - 3) = 22.5 sq. ft.

Now, the are of the right side triangle = \frac{1}{2} \times 10 \times 12 = 60 sq. ft.

Now, the area of non-shaded region = Total are of the two right triangles  

= (60 + 22.5) = 82.5 sq. ft.

Therefore, the area of the shaded region is = (Area of the total rectangle - Area of the non-shaded region)

= [(10 + 5) × 12] - 82.5

= 97.5 sq. ft. (Answer)

Wittaler [7]3 years ago
3 0

Answer:

<em>97.5 sq. ft.</em>

Step-by-step explanation:

Im presuming the question asks to find area of the shaded region.

First of all, the total figure is a rectangle. We can write an expression(in words) for the shaded area.

<em>Shaded Area = Area of Rectangle - Area of Small Triangle(White) - Area of Large Triangle(White)</em>

Now, we find respective areas.

Area of rectangle:

length * width = (5+10) * (12) = 15 * 12 = 180

Area of Small Triangle (white):

A = (1/2) * base * height = (1/2) * 5 * (12-3) = (1/2) * 5 * 9 = 22.5

Area of Large Triangle (white):

A = (1/2) * base * height = (1/2) * 10 * (12) = 60

Now, we find area of shaded region:

<em>Shaded Area = Area of Rectangle - Area of Small Triangle(White) - Area of Large Triangle(White)</em>

<em>Shaded Area = 180 - 22.5 - 60 = 97.5 sq. ft.</em>

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1. What is the least common multiple of both 9 and 12?
Kisachek [45]
1. the least common multiple of 9 and 12.
  9 18 27 36
  12 24 36
  36
2. GCF of 9 and 12
  1 3 9
  1 2 3 4 6 12
  3
7 0
3 years ago
I need help on Part B and C on question 2
kap26 [50]

Answer:

The answer to your question is

Part A. Archimedes grades 6 1/4 tests per day

Part B. 8 19/32 days

Part C.  6  26/29 days

Step-by-step explanation:

Part A

Total time = 6 2/5 days

Number of tests = 40 tests

Process

1.- Convert the mixed fraction to improper fraction

                 6 2/5 = (30 + 2) / 5 = 32/5

2.- Divide 40 by 32/5

                40/1  / 32/5   = (40 x 5) / (32 x 1)

                                      = 200 / 32

Simplify

                                        100 / 16 = 50/8 = 25/4

3.- Convert 25/4 to mixed fractions

                                       6      

                                4   25            

                                        1

                   25/4 = 6 1/4

Archimedes grade 6 1/4 tests per day

Part B

15 more tests

Total time = 32/5

Total tests = 40 + 15 = 55

Process

1.- Divide 55 by 32/5

                                  55 / 1  /  32 /5    = (55 x 5) / (32 x 1)

                                                             = 275 / 32

                                                             

2.- Convert 275/32 to a mixed fraction

                                                 8                                  

                                      32   275

                                             256

                                                19

Result    8 19/32 days

Part C

1.- Divide 55 by 7.25

                                         50 / 7.25     =    5000 / 725

                                                  6

                                725    5000

                                        - 4350

                                            650    

Result 6 650/750 = 6  26/29

3 0
3 years ago
A. (-10,-10)<br> B. (-1 /12, 9 1/2) <br> C. (1/2, 4) <br> D. (1,8)
lara31 [8.8K]

Answer:

I think The answer is ( -10, 10 )

5 0
2 years ago
Please help! I have more questions also.
Flura [38]
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3 0
3 years ago
Which graph represents the function f (x) = StartFraction 2 Over x minus 1 EndFraction + 4?
Pepsi [2]

Answer:

On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4

Step-by-step explanation:

The given function is presented as follows;

f(x) = \dfrac{2}{x - 1} + 4

From the given function, we have;

When x = 1, the denominator of the fraction, \dfrac{2}{x - 1}, which is (x - 1) = 0, and the function becomes, \dfrac{2}{1 - 1} + 4 = \dfrac{2}{0} + 4 = \infty + 4 = \infty therefore, the function in undefined at x = 1, and the line x = 1 is a vertical asymptote

Also we have that in the given function, as <em>x</em> increases, the fraction \dfrac{2}{x - 1} tends to 0, therefore as x increases, we have;

\lim_  {x \to \infty}  \dfrac{2}{(x - 1)} \to 0, and \  \dfrac{2}{(x - 1)}  + 4 \to 4

Therefore, as x increases, f(x) → 4, and 4 is a horizontal asymptote of the function, forming a curve that opens up and to the right in quadrant 1

When -∞ < x < 1, we also have that as <em>x</em> becomes more negative, f(x) → 4. When x = 0, \dfrac{2}{0 - 1} + 4 = 2. When <em>x</em> approaches 1 from the left, f(x) tends to -∞, forming a curve that opens down and to the left

Therefore, the correct option is on a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 4.

5 0
3 years ago
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