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4 years ago

Find the area of the following shape. You must show all work to receive credit.

Mathematics

1 answer:

Juliette [100K]4 years ago

7 0

Answer:

51.5 square units

Step-by-step explanation:

To find the area of an irregular shape, one method that can be used is to divide the area into simple shapes. Then you calculate the area of each simple shape using a known formula. Finally, you add all the areas together.

Let's divide the area into 2 trapezoids and 1 triangle.

First trapezoid at the bottom:

Draw a segment from point F to (11, -6), which is 2 units above B.

Below this segment, there is a trapezoid with base lengths: 11 and 6, and height 2.

Area of first trapezoid = (b1 + b2)h/2 = (11 + 6)(2)/2 = 17

Second trapezoid in the middle:

Draw a segment from point E through point D and ending at point (11, -3).

Below this segment, there is a trapezoid with base lengths: 11 and 7, and height 3.

Area of second trapezoid = (b1 + b2)h/2 = (11 + 7)(3)/2 = 27

Triangle on top:

The segment from point D to point (11, -3) is the bottom side of a triangle whose base is 5 and whose height is 3.

Area of triangle = bh/2 = (5)(3)/2 = 7.5

Total Area = Area of first trapezoid + area of second trapezoid + area of triangle

Total Area = 17 + 27 + 7.5

Total Area = 51.5

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A player on a school baseball team reaches first base 7/10 of the time he is at bat. Out of 50 times at bat, how many times woul

riadik2000 [5.3K]

Answer:

Step-by-step explanation:

You can find it by multiplying the decimal of the fraction (In this case it's 0.7) by however many times it's happening (50).

0.7 x 50 = 35

3 0

3 years ago

Solve 2x2 + x − 4 = 0. x2 + x + = 0

Ksenya-84 [330]

Answer:

The Quadratic Polynomial is

2 x² +x -4=0

Using the Determinant method to find the roots of this equation

$x=\frac{-1\pm \sqrt{1^2-4*2*(-4)}}{2\times 2}\\\\ x=\frac{-1 \pm \sqrt{32}}{4}{\text{Using the formula}}x=\frac{-b\pm\sqrt{D}}{2a}, {\text{where}},D=b^2-4 ac$

For, the Quadratic equation , ax²+ b x+c=0

(b) x²+x=0

x × (x+1)=0

x=0 ∧ x+1=0

x=0 ∧ x= -1

You can look the problem in other way

the two Quadratic polynomials are

2 x²+x-4=0, ∧ x²+x=0

x²= -x

So, 2 x²+x-4=0,

→ -2 x+x-4=0

→ -x -4=0

→x= -4

∨

x² +x² +x-4=0

x²+0-4=0→→x²+x=0

→x²=4

x=√4

x=2 ∧ x=-2

As, you will put these values into the equation, you will find that these values does not satisfy both the equations.

So, there is no solution.

You can solve these two equation graphically also.

3 0

4 years ago

Read 2 more answers

Find the distance between the two given points. (-6, -5) and (2, 0)

Ipatiy [6.2K]

Answer:

The answer is

<h2> $\sqrt{41} \: \: or \: \: \: 6.403 \: \: \: units$ </h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{( {x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(-6, -5) and (2, 0)

The distance is

$d = \sqrt{ ({ - 6 - 2})^{2} + ({ - 5 - 0})^{2} } \\ = \sqrt{ ({ - 4})^{2} + ({ - 5})^{2} } \\ = \sqrt{16 + 25} \: \: \: \: \: \: \: \: \: \: \: \: \: \:$