All categories

3 years ago

Find the area of the trapezoid

Mathematics

2 answers:

jenyasd209 [6]3 years ago

6 0

Answer:

its D

Step-by-step explanation:

Marrrta [24]3 years ago

3 0

Count the number of Little boxes there are on each side Like I did in the picture below. Then separate the triangle and the rectangle so that you can find the area of the rectangle first which you can figure out with just doing base times height. Then find the area of the triangle which is base times height divided by 2. The area of the rectangle should be 20 and the area of the triangle should be 6. Then you just add the two together so the area of a trapezoid should be 26. Now this is the area of the trapezoid not the perimeter. Perimeter and area are different things to find the perimeter you just count the outside lines of the trapezoid so it would be 5+4+5+3 Plus the long side of the triangle which you find by using Pythagorean theorem (A squared plus B squared equals C squared) I also show this in the picture below.

You might be interested in

PLSSSSS ALGEBRA IS HAARDDD

mario62 [17]

Ikr like I’m failing lol

6 0

3 years ago

halloween hut ordered 80 boxes with 20 costumes in each box. they also ordered 12 boxes with 90 masks in each box. how many more

tangare [24]

80x20=1600 costumes
12x90=1080 masks
1600-1080=520
they ordered 520 more costumes than masks. hope this helps!

8 0

3 years ago

Help please 50 points!

uranmaximum [27]

Answer:

A and B

Step-by-step explanation:

we would like to solve the following equation:

$\rm\displaystyle 5 - 2x = \sqrt{ {2x}^{2} + x - 1 } - x$

to do so isolate -x to the left hand side and change its sign:

$\rm\displaystyle 5 - 2x + x = \sqrt{ {2x}^{2} + x - 1 }$

simplify addition:

$\rm\displaystyle 5 - x = \sqrt{ {2x}^{2} + x - 1 }$

square both sides:

$\rm\displaystyle (5 - x {)}^{2} = (\sqrt{ {2x}^{2} + x - 1 } {)}^{2}$

simplify square of the right hand side:

$\rm\displaystyle (5 - x {)}^{2} = {2x}^{2} + x - 1$

use (a-b)²=a²-2ab+b² to expand the left hand side:

$\rm\displaystyle {x}^{2} - 10x + 25= {2x}^{2} + x - 1$

swap the equation:

$\rm\displaystyle {2x}^{2} + x - 1 = {x}^{2} - 10x + 25$

isolate the right hand side expression to the left hand side and change every sign:

$\rm\displaystyle {2x}^{2} + x - 1 - {x}^{2} + 10x - 25 = 0$

simplify:

$\rm\displaystyle {x}^{2} + 11x - 26= 0$

rewrite the middle term as 13x-2x:

$\rm\displaystyle {x}^{2} + 13x - 2x - 26= 0$

factor out x:

$\rm\displaystyle x({x}^{} + 13)- 2x - 26= 0$

factor out -2:

$\rm\displaystyle x({x}^{} + 13)- 2(x + 13)= 0$

group:

$\rm\displaystyle (x- 2)(x + 13)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases}x- 2 = 0\\ x + 13= 0 \end{cases}$

solve for x:

$\displaystyle \begin{cases}x = 2\\ x = - 13 \end{cases}$

to check for extraneous solutions we can define the domain of equation recall that a square root of a function is always greater than or equal to 0 therefore

$\rm\displaystyle 5 - x \geq0$