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2 years ago

Please solve this problem (surface area). With work bc my teacher said I need to show work with my answer

Mathematics

1 answer:

gregori [183]2 years ago

5 0

There are two triangular faces and three rectangular faces.
The triangular faces are congruent, so they have the same area. We calculate one and we'll know both.
The three rectangles are different.

Triangular face (each)
Although you can hardly see in the colored figure, there is a small symbol on the right side showing that the 4 km side and the 3 km side of the triangles are perpendicular. That means those sides are the base and the height.
A = (1/2)bh = (1/2)(3 km)(4 km) = 6 km^2

Rectangular faces:
A = LW = 7 km * 3 km = 21 km^2
A = LW = 7 km * 5 km = 35 km^2
A = LW = 7 km * 4 km = 28 km^2

total surface area = 2 * 6 km^2 + 21 km^2 + 35 km^2 + 28 km^2

total surface area = 96 km^2

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3 years ago

Can someone help me turn these two word problems into algebraic expressions?

Dima020 [189]

1. N=x+3
S=2(x+3), or Sam=2x+6

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3 years ago

The shape below is a composite figure made up of two rectangular prisms, what is the volume of the figure?

gregori [183]

Answer:

11098 cm³

Step-by-step explanation:

Volume of top prism

V = 62 x 3 x (13 - 7)
V = 186 x 6
V = 1116 cm³

Volume of bottom prism

V = 62 x 23 x 7
V = 1426 x 7
V = 9982 cm³

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⇒ Total volume = 11098 cm³

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2 years ago

Read 2 more answers

Find the area of the shaded regions. Give your answer as a completely simplified exact value in terms of π (no approximations).

Elan Coil [88]

Answer:

(40/3)*pi

Step-by-step explanation:

lets first find the area of the little circle

A=pi*r^2

A=pi*3^2

A=9pi

Now, lets find the area of the big circle

A=pi*r^2

A=pi*(3+4)^2

A=pi*7^2

A=49pi

now lets subtract the area of the little circle from the area of the big circle

49pi-9pi=40pi, now we found the area of the "bagel)

lets find what portion of the bagel is the shaded region

120/360=1/3

now lets multiply the bagel area by the fraction

40pi*(1/3)=

(40/3)*pi or (40*pi)/3

4 0

3 years ago

3. Using techniques from Calculus, show directly that the maximum value of a 1-D Gaussian distribution occurs at the point x = μ

Vilka [71]

Answer:

For a scaler variable, the Gaussian distribution has a probability density function of

p(x |µ, σ² ) = N(x; µ, σ² ) = 1 / 2π× $e^{\frac{-(x-u)^{2}}{2s^{2} } }$

The term will have a maximum value at the top of the slope of the 1-D Gaussian distribution curve that is when exp(0) =1 or when x = µ

Step-by-step explanation:

Gaussian distributions have similar shape, with the mean controlling the location and the variance controls the dispersion

From the graph of the probability distribution function it is seen that the the peak is the point at which the slope = 0, where µ = 0 and σ² = 1 then solution for the peak = exponential function = 0 or x = µ

5 0

3 years ago