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The area of a trapezoid is the sum of the bases divided by 2, all multiplied by the height.

Here, the two bases are 12 and 20, and the height is 15.

Find \(\frac{12+20}{2}\times 15\), which is the area of the barn.

Finally, subtract the door size. I think you can do that.

You are very welcome!

:P

The quadratic is (z + 7)^2 - 4 = 0 ==> z^2 + 14z + 45 ==> a + b = 59

Answer: 72

Explanation:

ALWAYS DRAW A DIAGRAM

First find the measure of an interior angle of a decagon.

180-360/n, n=10, so it's 144 degrees.

There are trapezoids GHIJ and DEFG, where DE and FG and GH and IJ are congruent and equal the side length.

The angle HGJ is one angle of the trapezoid GHIJ, and we can find that angle is 36 degrees (remember to draw it out).

Same with angle FGD.

Therefore, angle DGJ is 144 (the interior angle) minus 36 twice, or 144-36*2=72 degrees.

a=72.

You are very welcome!

:P

Pythagorean Theorem      c^    =   a^2 + b^2

                                         13^2 = 12^2 + b^2       solve for 'b'  

Angle Q and angle P are equal (see the equal    opposite sides?) = 3x+3

All three angles sum to 180 degrees

3x+3     +     3x+ 3    +  5x-2   = 180 ^o           Just solve for 'x'    thn angle Q =   3x + 3

11x    +4    = 180

11x + 176

x = 16     so angle Q = 3(11)+3 = 36^o

Break the shape up into parts. I break it up into a right triangle at the top, a rectangle in the middle, a tiny rectangle on the left, and a medium rectangle on the bottom.

Area of triangle: 17*5/2=42.5 mi

Can you do the rest?

You are very welcome!

:P

also to add on, I think we are not supposed to answer cheaters who just want answers to quizzes that they know they are going to flunk??!!!!!!!!! CoolStuffYT, its okay, I am grumpy:(, you are nice:)

A full circle is 2pi radians. 1.25 pi radians goes 5/8 around the circle.

I think you can solve from there - just find the area of the circle and multiply by 5/8.

You are very welcome!

:P

Thanks a lot Geno, I clearly understand the problem now. :)

Let's simplify the equation, getting a quadratic:

    (2x + 7)(x - 5)  =  -43 +jx

      2x² - 3x - 35  =  -43 + jx

       2x² - 3x + 8  =  jx

 2x² - 3x - jx + 8  =  0

2x² - (3 + j)x + 8  =  0

This will have only one solution when the discriminant is zero.

      a = 2     b = -(3 + j)     c  = 8

b² - 4·a·c  =  0     --->       [ -(3 + j) ]² - 4·2·8  =  0     --->     9 + 6j + j² - 64  =  0

                                                                                  --->     j² + 6j - 55  =  0

                                                                                  --->     (j + 11)(j - 5)  =  0

                                                                                  --->     j  =  -11     or    j  =  5

The possible values of j are either  -11  or  5

Each number is a multiple of 3, or 1 more than a multiple of 3, or one less than a multiple of 3.  These number we will call x's, y's, and z's.

If we add x + y + z, then we get x, because y and z will even each other out, since y is 1 more than multiple of 3 and z is one less.

Because of this, we can conclude that we have to arrange the numbers in a repeating pattern, like x, y, z, x, y, z, ... or z, y, x, z, y, x, ...

In this set 1, 2, 3, ..., 9, there are 3 x's, 3 y's, and 3 z's.

To fill up a pattern, there are 3 choices for the first x, y, z, 2 choices for second, then 1 choice.

3^3 * 2^3 * 1^3 = 27*8*1 = 216

This is equal for both patterns, so there are 216 * 2 = 432 ways to arrange the numbers.

Hi guest!

First off, we can start by subtracting 7 from both sides. We then get \(-2|7-9c|=-22\)

Then, we can divide both sides by -2. Now, we have \(|7-9c|=11\).

Since there are the absolute value brackets, there are 2 different scenarios:

\(7-9c=11\), AND \(7-9c=-11\). Now we just solve for both equations.

\(7-9c=11\)

\(-9c=4\)

\(c=-\frac{4}{9}\)

AND

\(7-9c=-11\)

\(-9c=-18\)

\(c=2\)

So, the final answer is \(\boxed{c=2 \text{ AND } c=-\frac{4}{9}}\)

I hope this helped you, guest!

If you have any questions, don't hesitate to ask!

:)

The outside circle has a radius of 10 cm; therefore, the area of the whole logo is:  pi·10²  or  100pi

The outside of the blue region is a circle of radius 7 cm; therefore its area is:  pi·7²  or  49pi

The inside of the blue region is a circle of radius 4 cm; therefore its area is:  pi·4²  or  16pi

The blue area is the outside region minus the inside region:  49pi - 16pi   =  33pi

The percentage of the logo that is blue is:  33pi / 100pi  =  33/100  =  0.33  =  33%

angle(VPL)  =   ½·arc(VKP)  =  ½·148^o  =  74^o

angle(JPV) + angle(VPL)  =  180^o     --->     angle(JPV) + 74^o  =  180^o​     --->     angle(JPV  =  106^o

(6, 7, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 67, 70, 71, 74, 75, 78, 79, 82, 83, 86, 87, 90, 91, 94, 95, 98) =47 terms

it is wrong?????

1/L + 1/C =1/2,
1/C + 1/M=3/5,
1/L+1/C+1/M =1,
1/L + 1/M =1/H, solve for C,H, L,M
C =10 hours for Curly
L =2.5 hours for Larry
M =2 hours for Moe
1/2.5 + 1/2 =1/H
H =1 1/9 - hours for Larry and Moe

The solution it gives me is:

There are \({10 \choose 2} =45\) ways to choose two of the points to connect with a segment, so there are \(\boxed{45}\) segments drawn.

wow! that's right! thanks Melody!

THX 

Yes but I would like to know how to solve this with algebra

Setting the denominator to zero will break up the domain into 5 parts.

2x² + x - 6  =  0     --->     (2x - 3)(x + 2)  =  0     --->     x = 1.5  and  x = -2.

The five parts are:  1)  the region below x = -2

                                2)  x = -2                                                              <--- at this point there is a break in the graph

                                3)  the region between x = -2 and x = 1.5

                                4)  x = 1.5                                                            <--- at this point there is a break in the graph

                                5)  the region above x = 1.5

Choose a number in the region below x = -2  ---   any number, I chose -5.  Place this number into the problem to see

if the answer is above 0 or below 0. It's above 0, so this region "works".

Choose a number in the region between x = -2 and x = 1.5  ---   any number, I chose 0. Place this number into the

problem to see if the answer is above 0 or below 0. It's below 0, so this region "doesn't work".

Choose a number in the region above x = 1.5  ---   any number, I chose 5.  Place this number into the problem to see

if the answer is above 0 or below 0. It's above 0, so this region "works".

So, the answer is:  (-infinity, -2)  union  (1.5, infinity)