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Anna35 [415]
3 years ago
10

1 Elon Musk has hired you to refabricate his private plane with carbon fibers. This is your chance for a lucrative career in the

starship industry and to set your parents up for a very comfortable life and their golden years.
Break the blue parts of the plane into at least four shapes. For each shape, provide the following 1) shapes drawn on plane. 2) formula used, 3) all work calculated the area, 4) area of the shape, 5) sum of all blue Show all work on the graph below

Mathematics
1 answer:
Allushta [10]3 years ago
5 0

The area of the blue parts of the plane comprises of four trapezoids,

from which each area can be calculated.

Response:

1) The drawings of the shapes are attached

2) \hspace{0.3 cm}Area \ of \ a \ trapezoid \ A = \dfrac{a + b}{2} \times h

3) The calculation for the area of each shape are;

  • Shape \ 1;\dfrac{2.2 + 1.6}{2} \times 4.6 = 8.74
  • Shape \ 2;\dfrac{2.2 + 1.6}{2} \times 4.6 = 8.74
  • Shape \ 3; \dfrac{0.9 + 1.7}{2} \times 1.6 = 2.08
  • Shape \ 4; \dfrac{0.9 + 1.7}{2} \times 1.6 = 2.08

4) Area of each shape is as follows;

  • Shape 1; Area = 8.74 square units
  • Shape 2; Area = 8.74 square units
  • Shape 3; Area = 2.06 square units
  • Shape 1; Area = 2.08 square units

5) Sum of all blue areas is 21.64 square units

<h3>Which method is used to break the blue parts?</h3>

The type of shape the blue parts break into are trapezoids

Area \ of \ a \ trapezoid \ A = \mathbf{\dfrac{a + b}{2} \times h}

Where;

<em>a</em>, and <em>b</em> are the lengths of the parallel sides;

h = The height of the trapezoid

The coordinates of the four shapes are;

  • Shape 1 Coordinates; (0.6, 0), (-1.6, 0), (-0.4, 4.6), and (1.2, 4.6)

1) Please find attached the drawing of shape 1 created with MS Excel.

2) The formula for the area trapezoid formed by shape 1 is; \dfrac{a + b}{2} \times h

3) The area of the shape is, A = \dfrac{2.2 + 1.6}{2} \times 4.6 = \mathbf{8.74}

4) Area of shape 1 is <u>8.74 square units</u>

  • Shape 2 Coordinates: (0.6, 0), (-1.6, 0), (0.4, -4.6), and (1.2, -4.6)

1) The drawing of shape 2 created with MS Excel is attached

2) The formula for the area trapezoid formed by shape 2 is; \dfrac{a + b}{2} \times h

3) The area of the shape 2 is, A = \dfrac{2.2 + 1.6}{2} \times 4.6 =\mathbf{ 8.74}

4) Area of shape 2 is <u>8.74 square units</u>

  • Shape 3 Coordinates: (4, 0), (4.6, 1.6), (5.5, 1.6), (5.7, 0)

1) The drawing of shape 3 created with MS Excel is attached

2) The formula for the area trapezoid formed by shape 2 is; \dfrac{a + b}{2} \times h

3) The area of the shape 3 is, A = \dfrac{0.9 + 1.7}{2} \times 1.6 = \mathbf{ 2.08}

4) Area of shape 3 is <u>2.08 square units</u>

Coordinates of shape 4; (4, 0), (4.6, -1.6), (5.5, -1.6), (5.7, 0)

1) The drawing of shape 4 created with MS Excel is attached

2) The formula for the area trapezoid formed by shape 2 is; \dfrac{a + b}{2} \times h

3) The area of the shape 4 is, A = \dfrac{0.9 + 1.7}{2} \times 1.6 = \mathbf{2.08}

4) Area of shape 4 is<u> 2.08 square units</u>

  • 5) The sum of the area of the blue shapes = 2 × 8.74 + 2 × 2.08 = <u>21.64</u>

Learn more about composite figures here:

brainly.com/question/8971404

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The figure is a right triangle so we can use the Pythagorean theorem to find the third side AKA the hypotenuse.

Pythagorean theorem = a² + b² = c² where A and B are the legs of the triangle and C is the hypotenuse.

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13. Which equation represents the line below?
Nonamiya [84]

Answer:

y = -2/3x + 3

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First, we need to calculate the slope of the line. We can do this by using the slope formula using two points:

(y₂ - y₁) / (x₂ - x₁)

If we take the points (-3,5) and (3,1), then: y₂ = 1; y₁ = 5; x₂ = 3; x₁ = -3

Next, we can substitute in values in the equation above to get (1 -5)/(3+3)

<em>** We got 3+3 because it was 3 - (-3), where two negatives equal a positive**</em>

Then, we get, -4/6, which simplifies to -2/3. Our slope is -2/3

After that, we need to find the y-intercept, which is where the line intercepts the y-axis. Here, that is at (0,3).

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Though I don't think that answer choice is there...

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A tank contains 1080 L of pure water. Solution that contains 0.07 kg of sugar per liter enters the tank at the rate 7 L/min, and
allsm [11]

(a) Let A(t) denote the amount of sugar in the tank at time t. The tank starts with only pure water, so \boxed{A(0)=0}.

(b) Sugar flows in at a rate of

(0.07 kg/L) * (7 L/min) = 0.49 kg/min = 49/100 kg/min

and flows out at a rate of

(<em>A(t)</em>/1080 kg/L) * (7 L/min) = 7<em>A(t)</em>/1080 kg/min

so that the net rate of change of A(t) is governed by the ODE,

\dfrac{\mathrm dA(t)}[\mathrm dt}=\dfrac{49}{100}-\dfrac{7A(t)}{1080}

or

A'(t)+\dfrac7{1080}A(t)=\dfrac{49}{100}

Multiply both sides by the integrating factor e^{7t/1080} to condense the left side into the derivative of a product:

e^{\frac{7t}{1080}}A'(t)+\dfrac7{1080}e^{\frac{7t}{1080}}A(t)=\dfrac{49}{100}e^{\frac{7t}{1080}}

\left(e^{\frac{7t}{1080}}A(t)\right)'=\dfrac{49}{100}e^{\frac{7t}{1080}}

Integrate both sides:

e^{\frac{7t}{1080}}A(t)=\displaystyle\frac{49}{100}\int e^{\frac{7t}{1080}}\,\mathrm dt

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Solve for A(t):

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Given that A(0)=0, we find

0=\dfrac{378}5+C\implies C=-\dfrac{378}5

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\boxed{A(t)=\dfrac{378}5\left(1-e^{-\frac{7t}{1080}}\right)}

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\displaystyle\lim_{t\to\infty}A(t)=\frac{378}5

or 75.6 kg of sugar.

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