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Dvinal [7]
3 years ago
8

Can someone solve this! Need it done soon

Mathematics
1 answer:
Airida [17]3 years ago
7 0

Answer:

117.5 ft²

Step-by-step explanation:

To find the area of the shaded area, we can find the area of the square, then subtract the areas of the two semicircles.

First, we will find the area of the given square, by using the formula lw, multiplying the length and width. The dimensions of this square are 14×14.

14 · 14 = 196

The area of the square is 196 ft².

We can now find the area of the two congruent half-circles. Since they are identical, we can simply find the area of one circle if it was whole. To find the area of a circle, we'll use the formula A=\pi r^2. With some simple deduction, we can see that the diameter of the circle is 10 ft, so the radius would be 5 ft long. Plug our values into the formula.

A = \pi5²

We will use 3.14 for \pi.

A = 78.5

The area of both the semicircles is 78.5 ft².

Now, we can subtract.

196 - 78.5 = 117.5

The area of the figure is 117.5 ft².

Good luck ^^

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The weight Wkg of a metal bar varies jointly
maria [59]

Answer:

d = \sqrt{\frac{216W}{35L} }

Step-by-step explanation:

Given that W varies jointly as L and d² then the equation relating them is

W = kLd² ← k is the constant of variation

To find k use the condition W = 140 when d = 4 and L = 54, thus

140 = k × 54 × 4² = 864k ( divide both sides by 864 )

\frac{140}{864} = k , that is

k = \frac{35}{216}

W = \frac{35}{216} Ld² ← equation of variation

Multiply both sides by 216

216W = 35Ld² ( divide both sides by 35L )

\frac{216W}{35L} = d² ( take the square root of both sides )

d = \sqrt{\frac{216W}{35L} }

6 0
3 years ago
Are the figures below similar? Why or why not? Determine whether the triangles shown in the image are similar.
Vika [28.1K]

Answer:

B) no, because the corresponding angles are not congruent

Step-by-step explanation:

Similar triangles must have proportional sides, as well as congruent corresponding angles. In this instance, we can see that the angles are not <em>congruent</em>, and so there is no need to solve for proportion.

~

3 0
3 years ago
Read 2 more answers
Hey can you please help me posted picture of question
jenyasd209 [6]

The equation of parabola is x=ay^2. If a is positive and y^2 is always greater than zero or equal to zero, then x is also greater or equal to zero. This means that parabola is determined for non-negative x and for all real y.

Tha canonical equation of parabola is y^2=2px, where p>0. The branches of this parabola go up in positive y-direction. When you change x to y and y to x, then the branches of parabola go in positive x-direction, that is right.

Answer: correct choice is A.

6 0
3 years ago
Triangle PQR is transformed to similar triangle P′Q′R′ below:
morpeh [17]
1 over 3 should be the answer I think
6 0
3 years ago
Solving Exponential and Logarithmic Equations In Exercise, solve for x.<br> In 2x - In(3x - 1) = 0
LenKa [72]

Answer:

\frac{-1}{3x^2-x}

Step-by-step explanation:

  1. If f(x) is in th form of f(x)=g(x)-h(x) then f'(x)=g'(x) - h'(x)
  2. When f(x)=z(g(x)) then f'(x)= z'(g(x))g'(x) (called as chain rule)

<u>using these information</u>:

g(x)=ln2x then g'(x)=\frac{(2x)'}{2x} =\frac{2}{2x}=\frac{1}{x}

h(x)=In(3x - 1) then h'(x)=\frac{(3x-1)'}{3x-1} =\frac{3}{3x-1}f'(x)=g'(x) - h'(x) =[tex]\frac{1}{x} - \frac{3}{3x-1} =\frac{-1}{3x^2-x}

7 0
4 years ago
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