All categories

3 years ago

What is the area of the following trapezoid in square feet?

Mathematics

2 answers:

77julia77 [94]3 years ago

7 0

Answer:

135 ft^2 (first answer choice)

Step-by-step explanation:

Convert the length of the 7 yd base to feet: 21 ft.

Then the average length of the trapezoid is (15 ft + 21 ft)/2 = 36/2 ft = 18 ft.

The area of the trapezoid is then (average length)(height), or

A = (18 ft)(7.5 ft) = 135 ft^2 (first answer choice)

tangare [24]3 years ago

4 0

Answer:

82.5 ft²

Step-by-step explanation:

The area (A) of a trapezoid is calculated as

A = $\frac{1}{2}$ h (b₁ + b₂)

where h is the perpendicular height and b₁, b₂ the parallel bases

Here h = 7.5, b₁ = 7 and b₂ = 15 , then

A = $\frac{1}{2}$ × 7.5 × (7 + 15) = 0.5 × 7.5 × 22 = 82.5 ft²

You might be interested in

Tisha and her academic team are working to go to state finals. They must have a certain number of points to advance. They have h

kolbaska11 [484]

Answer:

$b=\frac{T-a-c-d}{3}$

b = (T - a - c - d) / 3

Step-by-step explanation:

Let T be the total number of points required to advance.

a, c and d are points scored in the local matches, and b is the number of points scored in the district match. If b is worth 3 times as much as the other matches, the total number of points is given by:

$T=a+c+d+3b$

Isolate b in order to find out how many points they need in the district match:

$T=a+c+d+3b\\3b=T-a-c-d\\b=\frac{T-a-c-d}{3}$

They need to score (T - a - c - d)/3, in the district match in order to win.

4 0

3 years ago

Answer 44 and 48<br>only 44 and 48

joja [24]

I know the correct answers but I need you to answer my math questions first

7 0

3 years ago

6 millimeters = yards

sertanlavr [38]

6 millimeters equals 0.00656168 yards.

7 0

3 years ago

7 yards and 2 feet is equal to how many feet

alisha [4.7K]

<span>7 yards 2 feet = 23 feet
</span>

3 0

3 years ago

Read 2 more answers

Consider the function g(x) = (x-e)^3e^-(x-e). Find all critical points and points of inflection (x, g(x)) of the function g.

Elden [556K]

Answer:

The answer is " $cirtical\ points \ (x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})$ "

Step-by-step explanation:

Given:

$g(x) = (x-e)^3e^{-(x-e)}$

Find critical points:

$g(x) = (x-e)^3e^{(e-x)}$

differentiate the value with respect of x:

$\to g'(x)= (x-e)^3 \frac{d}{dx}e^{e-r} +e^{e-r} \frac{d}{dx}(x-e)^3=(x-e)^2 e^{(e-x)} [-x+e+3]$

critical points $g'(x)=0$

$\to (x-e)^2 e^{(e-x)} [e+3-x]=0\\\\\to e^{(e-x)}\neq 0 \\\\\to (x-e)^2=0\\\\ \to [e+3-x]=0\\\\\to x=e\\\\\to x=e+3\\\\\to x= e,e+3$

So,

The critical points of $(x,g(x))\equiv (e,0),(e+3,\frac{27}{e^3})$

7 0

3 years ago