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Maksim231197 [3]
2 years ago
12

What is the area of this figure? Enter your answer in the box. ___m²

Mathematics
1 answer:
ra1l [238]2 years ago
6 0
Well first you got figure out the formulas for the 2 triangles and the trapezoid.

The triangles formula for area is: A=b*h/2

The trapezoids formula for area is: A=1/2(b1+b2)*h

After you found the formulas you got to plug in the numbers.

For the first triangle it’s: A=9*5/2

The second triangle it’s: A=8*8/2

The trapezoid it’s: A=1/2(9+8)*5

After you plug in the numbers you have to solve the areas.

So the first triangle is: 9*5/2=9*5=45/2=22.5

So the area of the first triangle is 22.5

Then you have to solve the second triangle.

The second triangle: 8*8/2=8*8=64/2=32

So the area of the second triangle is 32

After solving the second triangle you have to solve the trapezoid.

Trapezoid: 1/2(9+8)*5=9+8=17=1/2(17)*5=17*5=85/2=
42.5

The area of the trapezoid is 42.5

After solving all of the shapes you have to add up all of the areas to get your full area.

22.5+32+42.5=22.5+42.5=65+32=97

So the area of the whole thing is 97

Hope this helps you!!!! Have a nice day!!!!
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Find the slope and the y-intercept of the line. 4x-5y=5
Evgen [1.6K]

Answer:

Step-by-step explanation:

4x - 5y= 5

-4x. -4x

-5y = -4x + 5

divide all by -5

y= 4/5x -1

slope is 4/5

y intercept is -1

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2 years ago
Estimate the quotient for 469.4÷62​
Alex Ar [27]

Answer:

469.4/62=7.57

Step-by-step explanation:

its the correct answer

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Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices
Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt

Where P, Q are scalar functions

We want to compute

\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths \large C_1,C_2,C_3,C_4 which represents the sides of the rectangle.

\large C_1 is the line segment from (0,0) to (5,0)

\large C_2 is the line segment from (5,0) to (5,1)

\large C_3 is the line segment from (5,1) to (0,1)

\large C_4 is the line segment from (0,1) to (0,0)

Then

\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize \large C_1 as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

\large \displaystyle\int_{C_1}xydx+x^2dy=0

 Now the second integral. We parametrize \large C_2 as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25

The third integral. We parametrize \large C_3 as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2

The fourth integral. We parametrize \large C_4 as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

\large \displaystyle\int_{C_4}xydx+x^2dy=0

So

\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2

Now, let us compute the value using Green's theorem.

According with this theorem

\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx

where A is the interior of the rectangle.

so A={(x,y) |  0≤ x≤ 5,  0≤ y≤ 1}

We have

\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x

so

\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2

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solniwko [45]

Answer:

-9.85

Step-by-step explanation:

-15.5 - 4.2 = -19.7

Negatives work like positives

19.7 / 2 = 9.85

Turn that into a negative

-9.85

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2 years ago
-8+(-4)+(-2)+(-1)+(-1/2)+.... what is the sum of the first 7 terms of the geometric series
likoan [24]
Your answer is 15.875

-8+(-4)+(-2)+(-1)+(-1/2)+(-1/4)+(-1/8)
you half the previous vaule to find the next value

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