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Pavel [41]
2 years ago
12

What is the area of this quadrilateral?

Mathematics
1 answer:
soldier1979 [14.2K]2 years ago
4 0

Answer:

The area is 40 square feet

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100 points to whoever helps the fastest.
jonny [76]

Answer:

cos theta = 8/17

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

cos theta = adjacent side/ hypotenuse

cos theta = 8/17

6 0
3 years ago
Read 2 more answers
1 - 4 + (x + 6 * 1) when 'x' equals 25
ikadub [295]
The answer is 28. you start with the parenthesis and do 25+6 which is 31x1 is 31. then you do 1-4 and get -3. then add -3 to 31 and get 28
8 0
3 years ago
How many times does the graph of the function below intersect or touch the x-axis? y=-3x2 + x+4 A. 2 B. 1 C. 3 D. 0​
galben [10]

Answer:

this is your ans.. i hope it is helpfull

6 0
3 years ago
A force of 3 pounds is required to hold a spring stretched 0.6 feet beyond its natural length. how much work (in foot-pounds) is
ioda

The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

  • Force (F) = 3 pounds
  • Extension (e) = 0.6 feet
  • Work done (Wd) =?

<h3>How to determine the spring constant</h3>
  • Force (F) = 3 pounds
  • Extension (e) = 0.6 feet
  • Spring constant (K) =?

F = Ke

Divide both sides by e

K = F/ e

K = 3 / 0.6

K = 5 pound/foot

Thus, the spring constant of the spring is 5 pound/foot

<h3>How to determine the work done</h3>
  • Spring constant (K) = 5 pound/foot
  • Extention (e) = 0.7 feet
  • Work done (Wd) =?

Wd = ½Ke²

Wd = ½ × 5 × 0.7²

Wd = 2.5 × 0.49

Wd = 1.23 foot-pound

Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound

Learn more about spring constant:

brainly.com/question/9199238

#SPJ1

5 0
1 year ago
Find the solution to the system of equations below.<br> 2y = 2x - 8<br> 2x + y = 5
ladessa [460]

Answer:

The solution to the system of equations is:

  • (x, y) = (3, -1)

The graph is attached below.

Step-by-step explanation:

Given the system of equations

2y = 2x - 8

2x + y = 5

Let us solve the system of equations using the elimination method

\begin{bmatrix}2y=2x-8\\ 2x+y=5\end{bmatrix}

Arrange equation variables for elimination

\begin{bmatrix}2y-2x=-8\\ y+2x=5\end{bmatrix}

Multiply y + 2x = 5 by 2:  2y + 4x = 10

\begin{bmatrix}2y-2x=-8\\ 2y+4x=10\end{bmatrix}

subtracting the equations

2y+4x=10

-

\underline{2y-2x=-8}

6x=18

solve 6x = 18 for x

6x=18

Divide both sides by 6

\frac{6x}{6}=\frac{18}{6}

simplify

x=3

For 2y - 2x = -8  plug in x = 3

2y-2\cdot \:3=-8

2y-6=-8

Add 6 to both sides

2y-6+6=-8+6

Simplify

2y=-2

Divide both sides by 2

\frac{2y}{2}=\frac{-2}{2}

Simplify

y=-1

Therefore, the solution to the system of equations is:

  • (x, y) = (3, -1)

The graph is attached below.

8 0
3 years ago
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