All categories

2 years ago

Find the area of this composite shape. please help

Mathematics

1 answer:

VLD [36.1K]2 years ago

3 0

Answer:

triangle on the left: 6 sq ft. triangle on the right: 28 sq ft. rectangle: 128. altogether, 162 sq ft. the trapezoid is about 61.6.

Step-by-step explanation:

for the rectangle the formula is bxh (base x height). for a triangle, it is bxh/2, or base times height divided by 2. for a trapezoid, it is b1+b2 x h / 2, or base 1 (bottom) plus base 2 (top) times height, take all of that and divide by 2.

You might be interested in

Surface area of a cylinder with a radius of 7cm and a height of 12

Keith_Richards [23]

Answer:

CSA = 2 * pi * r * h

Step-by-step explanation:

CSA = 2 * 22 * 7 * 12 / 7

= 2 * 22 * 12

528 cm ^3

7 0

3 years ago

Read 2 more answers

Which statement best describes the function represented by the graph?

blondinia [14]

Answer:

The second one decreasing on the interval (-infinity, 0) where it becomes constant

4 0

3 years ago

Volume is always measured in square units. True False

bogdanovich [222]

Hello,

The answer is option B "false".

Reason:

The answer is false because the volume can be measure in anyway for example: feet, yards.

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit

4 0

3 years ago

Read 2 more answers

Consider the three points ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 ) . Let ¯ x be the average x-coordinate of these points, and let ¯ y

loris [4]

Answer:

$m=\dfrac{3}{2}$

Step-by-step explanation:

Given points are: ( 1 , 3 ) , ( 2 , 3 ) and ( 3 , 6 )

The average of x-coordinate will be:

$\overline{x} = \dfrac{x_1+x_2+x_3}{\text{number of points}}$

1) Finding $(\overline{x},\overline{y})$

Average of the x coordinates:

$\overline{x} = \dfrac{1+2+3}{3}$

$\overline{x} = 2$

Average of the y coordinates:

similarly for y

$\overline{y} = \dfrac{3+3+6}{3}$

$\overline{y} = 4$

2) Finding the line through $(\overline{x},\overline{y})$ with slope m.

Given a point and a slope, the equation of a line can be found using:

$(y-y_1)=m(x-x_1)$

in our case this will be

$(y-\overline{y})=m(x-\overline{x})$

$(y-4)=m(x-2)$

$y=mx-2m+4$

this is our equation of the line!

3) Find the squared vertical distances between this line and the three points.

So what we up till now is a line, and three points. We need to find how much further away (only in the y direction) each point is from the line.

Distance from point (1,3)

We know that when x=1, y=3 for the point. But we need to find what does y equal when x=1 for the line?

we'll go back to our equation of the line and use x=1.

$y=m(1)-2m+4$

$y=-m+4$

now we know the two points at x=1: (1,3) and (1,-m+4)

to find the vertical distance we'll subtract the y-coordinates of each point.

$d_1=3-(-m+4)$

$d_1=m-1$

finally, as asked, we'll square the distance

$(d_1)^2=(m-1)^2$

Distance from point (2,3)

we'll do the same as above here:

$y=m(2)-2m+4$

$y=4$

vertical distance between the two points: (2,3) and (2,4)

$d_2=3-4$

$d_2=-1$

squaring:

$(d_2)^2=1$

Distance from point (3,6)

$y=m(3)-2m+4$

$y=m+4$

vertical distance between the two points: (3,6) and (3,m+4)

$d_3=6-(m+4)$

$d_3=2-m$

squaring:

$(d_3)^2=(2-m)^2$

3) Add up all the squared distances, we'll call this value R.

$R=(d_1)^2+(d_2)^2+(d_3)^2$

$R=(m-1)^2+4+(2-m)^2$

4) Find the value of m that makes R minimum.

Looking at the equation above, we can tell that R is a function of m:

$R(m)=(m-1)^2+4+(2-m)^2$

you can simplify this if you want to. What we're most concerned with is to find the minimum value of R at some value of m. To do that we'll need to derivate R with respect to m. (this is similar to finding the stationary point of a curve)

$\dfrac{d}{dm}\left(R(m)\right)=\dfrac{d}{dm}\left((m-1)^2+4+(2-m)^2\right)$