3. Find the area of the trapezoid.

Mathematics

1 answer:

gavmur [86]3 years ago

5 0

b 28

hope this helps u

follow me

mark as brainliest

thank me

You might be interested in

Find the coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4)

Eva8 [605]

The coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4) is $(\frac{-17}{5} ,\frac{2}{5} )$

Explanation :

the coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4)

Apply section formula to find coordinates of point Q

$(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n} )$

Ratio m:n is 2:3 and point R is (x1,y1) , point S is (x2,y2)

Substitute all the values inside the formula

$(\frac{2(2)+3(-7)}{2+3} , \frac{2(4)+3(-2)}{2+3} )\\\\(\frac{-17}{5} ,\frac{2}{5} )$

The coordinates of point Q is $(\frac{-17}{5} ,\frac{2}{5} )$

Learn more : brainly.com/question/13878373

7 0

3 years ago

Simplify the expression to a polynomial in standard form: (3x-2)(-x^2+8x+6) (3x−2)(−x 2 +8x+6)

photoshop1234 [79]

Step-by-step explanation:

use mathaway! it helps a lot

3 0

3 years ago

Help!!!1!!!!!!!!!!!!!!!!!!!!!!

ad-work [718]

Equation B is written in vertex form, which means you can read the vertex (extreme value) from the numbers in the equation.

Vertex form is
y = a(x -h)² + k
where the vertex (extreme point) is (h, k). Whether that is a maximum or a minimum depends on the sign of "a". When "a" is negative, the graph is a parabola that opens downward, so the vertex is a maximum.

Equation B reveals its extreme value without needing to be altered.
The extreme value of this equation is a maximum at the point (2, 5).