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zmey [24]
3 years ago
8

Find the surface area of this triangular prism.​

Mathematics
1 answer:
12345 [234]3 years ago
7 0

Answer:

Surface area of the triangular prism is 1380m^2

Step-by-step explanation:

The triangular prism have two triangular faces and three rectangular faces:

So, the Area of the Prism is :

       Area of the two triangular faces+ Area of the three rectangular faces

Surface area of the triangular prism :    

                                                 2(1/2*24*10)+(19*10)+(24*19)+(26*19)\\\\                                                       = 240+190+456+494\\\\                                                        =1380 m^2\\

So, the area of the triangular prism is 1380m^2

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Identify a possible first step using the elimination method to solve the system and then find the solution to the system. 3x - 5
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Answer:

see explanation

Step-by-step explanation:

Given the 2 equations

3x - 5y = - 2 → (1)

2x + y = 3 → (2)

Multiply (2) by 5 will eliminate y when added to (1), that is

10x + 5y = 15 → (3)

Add (1) and (3) term by term

(3x + 10x) + (- 5y + 5y) = (- 2 + 15)

13x = 13 ( divide both sides by 13 )

x = 1

Substitute x = 1 into (2) for corresponding value of y

2 + y = 3 ⇒ y = 3 - 2 = 1

Solution is (1, 1)

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3 years ago
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What is the rate of change seen in the graph below?
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Step-by-step explanation:

This flat horizontal line is Y=1 has a slope (rate of change) of zero.

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2 years ago
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Jason decided that he will sell his stocks if their value per share (x) goes below $5 or above $15. Which compound inequality re
Veronika [31]

Answer:

  B.  x < $5 or x > $15

Step-by-step explanation:

The "or" in the problem statement gives you a clue that answer choices C and D can be rejected.

If x is the stock value, then value per share "below $5" means x < $5. This corresponds with answer choice B. Of course, "above $15" means x > $15, also corresponding to answer choice B.

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Consider the equation and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”. For example, “(0, 3) R (0, 2)”
Leviafan [203]

Answer:

The equation determine a relation between x and y

x = ± \sqrt{1-(y-2)^{2}}

y = ± \sqrt{1-x^{2}}+2

The domain is 1 ≤ y ≤ 3

The domain is -1 ≤ x ≤ 1

The graphs of these two function are half circle with center (0 , 2)

All of the points on the circle that have distance 1 from point (0 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

 R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

 of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* <em>Lets solve the problem</em>

- The equation of a circle of center (h , k) and radius r is

  (x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒  (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

∴ The equation determine a relation between x and y

* Lets prove that x=g(y)

- To do that find x in terms of y by separate x in side and all other

  in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

∴ x² = 1 - (y - 2)²

- Take square root for both sides

∴ x = ± \sqrt{1-(y-2)^{2}}

∴ x = g(y)

* Lets prove that y=h(x)

- To do that find y in terms of x by separate y in side and all other

  in the other side

∵ x² + (y - 2)² = 1

- Subtract x² from both sides

∴ (y - 2)² = 1 - x²

- Take square root for both sides

∴ y - 2 = ± \sqrt{1-x^{2}}

- Add 2 for both sides

∴ y = ± \sqrt{1-x^{2}}+2

∴ y = h(x)

- In the function x = ± \sqrt{1-(y-2)^{2}}

∵ \sqrt{1-(y-2)^{2}} ≥ 0

∴ 1 - (y - 2)² ≥ 0

- Add (y - 2)² to both sides

∴ 1 ≥ (y - 2)²

- Take √ for both sides

∴ 1 ≥ y - 2 ≥ -1

- Add 2 for both sides

∴ 3 ≥ y ≥ 1

∴ The domain is 1 ≤ y ≤ 3

- In the function y = ± \sqrt{1-x^{2}}+2

∵ \sqrt{1-x^{2}} ≥ 0

∴ 1 - x² ≥ 0

- Add x² for both sides

∴ 1 ≥ x²

- Take √ for both sides

∴ 1 ≥ x ≥ -1

∴ The domain is -1 ≤ x ≤ 1

* The graphs of these two function are half circle with center (0 , 2)

* All of the points on the circle that have distance 1 from point (0 , 2)

8 0
3 years ago
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