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diamong [38]
3 years ago
9

Find the area of the trapezoid.

Mathematics
2 answers:
solong [7]3 years ago
5 0
Add top and bottom number, divide by 2 then multiply by the height:

18 +6 = 24

24/2 = 12

 12 * 12 = 144 square inches

Vadim26 [7]3 years ago
4 0
Area of the trapezium = 1/2(18 + 6) x 12
Area of the trapezium = 144 in²

------------------------------------------------
Answer: Area = 144 in²
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P(g)=12/(12+8)

P(g)=12/20

P(g)=3/5
7 0
3 years ago
PLEASE HELP 100 POINTS.
aalyn [17]

Wait I don’t understand what u need
3 0
3 years ago
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A container holds 4 quarts of lemonade. How much is this in cups?
Fantom [35]
1 quart = 4 cups.  So 4 x 4 will give you your answer, which is 16 cups.
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4 years ago
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Please Help with angles!
777dan777 [17]

Step-by-step explanation:

003:

To solve this, we'll be using the Pythagorean theorem, a^2+b^2=c^2

a^2+8^2=15^2\\a^2+64=225\\a^2=161\\a = \sqrt{161}

004:

For the next couple questions we'll be using a fun rule called SOHCAHTOA, which stands for Sin (Opposite/Hypotenuse) Cos (Adjacent/Hypotenuse) Tan (Opposite/Adjacent). From this we can see that Tan θ is 8/\sqrt{161}.

005:

From this rule again, we can tell that sin Ф (Opposite/Hypotenuse) is going to be \sqrt{161}/15.

Answer:

003: \sqrt{161}

004: 8/\sqrt{161}

005: \sqrt{161}/15

7 0
3 years ago
Use calculus to find the first order conditions. Use solver then to solve the first order conditions. Show your work for the fir
DaniilM [7]

The values of x and y are 70/22 and 30/22 respectively by using the first-order condition of differential calculus.

<h3>What is the first-order condition in differential calculus?</h3>

A first-order differential equation is represented by the equation \mathbf{ \dfrac{dy}{dx} =f (x,y) }with 2 variables x & y, including its function f(x,y) specified on a xy-plane.

Given that:

\mathbf{f(x,y) =-22x^2+22xy-11y^2+110x-40y-23}

Let us first differentiate the above equation with respect to x, we have:

\mathbf{\dfrac{\partial f(x,y) }{\partial x}  = -44x +22y -0+110-0-0=0}

\mathbf{\implies  -44x +22y+110=0}        (multiply by -1)

44x - 22y = 110    ------ (equation 1)

Now, differentiating with respect to y, we have:

\mathbf{\dfrac{\partial f(x,y) }{\partial y}  =0 +22x-22y +0-40-0=0}

\mathbf{\implies 22x-22y -40=0}

22x - 22y = 40      ----- (equation 2)

Now, we have a system of equations:

44x - 22y = 110

-                      ---- ( subtracting equation 2 from 1; elimination method)

<u> 22x - 22y = 40  </u>

<u>22x    + 0  = 70    </u>

<u />

x = 70/22

Replacing the value of x into equation (1), we have:

44x - 22y = 110

44(70/22) - 22y = 110

140 - 22y = 110

140 - 110 = 22y

30 = 22y

y = 30/22

Learn more about the first-order conditions in differential calculus here;

brainly.com/question/14528981

#SPJ1

4 0
2 years ago
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