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Nat2105 [25]
3 years ago
7

The right isosceles triangle shown is rotated about line k with the base forming perpendicular to k. The perimeter of the triang

le is 58 units. Which best describes the resulting three-dimensional shape?

Mathematics
1 answer:
kipiarov [429]3 years ago
4 0

Answer:

The new shape will be cone with a radius of 17 units, tilt height 24 and units of height k.

Step-by-step explanation:

The image related to the exercise is necessary, to be able to solve therefore the attached one.

We have the following information:

Perimeter of the triangle = 58 units

Hypotenuse = 24 units

We have that the other two sides are equal and have x units, therefore we have the following:

x + x + 24 = 58

2 * x = 58-24

2 * x = 34

x = 34/2

x = 17

Now the triangle rotates around line k , and then it will result in a cone, which is a three-dimensional shape.

Therefore, the new shape will be cone with a radius of 17 units, tilt height 24 and units of height k.

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The incorrect work of a student to solve an equation 2(y + 6) = 4y is shown below.
Lina20 [59]

Answer:

<h3>⇒ A. 2 should be distributed as 2y + 12, y = 6</h3>

Step-by-step explanation:

Here are the steps to solve this equation:

<u>Given:</u>

First step: 2(y+6)-4y

Use the distributive property.

<u>Distributive property:</u>

<h3>⇒ A(B+C)=AB+AC</h3>

2(y+6)

2*y=2y

2*6=12

2y+12

Isolate the term of y from one side of the equations.

2y+12=4y

Subtract by 12 from both sides.

2y+12-12=4y-12

Solve.

2y=4y-12

Then, you subtract by 4y from both sides.

2y-4y=4y-12-4y

Solve.

2y-4y=-2y

-2y=-12

Divide by -2 from both sides.

\sf{\dfrac{-2y}{-2}=\dfrac{-12}{-2}}

Solve.

Divide the numbers from left to right.

<u>Solutions:</u>

-12/-2=6

\Longrightarrow: \boxed{\sf{y=6}}

  • <u>Therefore, the correct answer is A. "2 should be distributed as 2y+12, y=6".</u>

I hope this helps. Let me know if you have any questions.

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Answer: yt

Step-by-step explanation:

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san4es73 [151]

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3 years ago
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ss7ja [257]

The area of the <em>irregular</em> quadrilateral ABCD is equal to 234 square centimeters. (Correct choice: C)

<h3>What is the area of the quadrilateral?</h3>

Herein we have a description of an <em>irregular</em> quadrilateral, whose area must be determined by adding the areas of minor quadrilaterals and triangles that are part of it. The area is now determined:

A = 0.5 · (24 cm) · (7 cm) + 0.5 · (15 cm) · (20 cm)

A = 234 cm²

The area of the <em>irregular</em> quadrilateral ABCD is equal to 234 square centimeters. (Correct choice: C)

<h3>Remark</h3>

The picture with the quadrilateral is missing and is included as attachment.

To learn more on quadrilaterals: brainly.com/question/13805601

#SPJ1

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