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nikklg [1K]
3 years ago
13

Which figure shows how a shape can be rotated about an axis to form a cylinder?

Mathematics
2 answers:
cluponka [151]3 years ago
4 0

Answer:

the bottom right

Step-by-step explanation:

ANTONII [103]3 years ago
4 0

Answer

the rectangle

Step-by-step explanation:

the cylinder folds like a circle but it is not a circle.

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3x+8-X=57-4<br> Is 7 a solution to the question
Diano4ka-milaya [45]

Answer:

7 is not a solution for x.

Step-by-step explanation:

To see if 7 is a solution for x, we will simply plug in the value for x and see if the left hand side is equal to the right hand side.

3x + 8 - x = 57 - 4

3(7) + 8 - (7) ?= 57 - 4

21 + 8 - 7 ?= 53

29 - 7 ?= 53

22 ?= 53  (( NO ))

Since 22 does not equal 53, 7 is not a solution for x.

Let's find the solution for x:

3x + 8 - x = 57 - 4

2x + 8 = 53

2x = 45

x = 22.5

Let's validate this solution for x:

3x + 8 - (x) = 57 - 4

3(22.5) + 8 - (22.5) ?= 57 - 4

67.5 + 8 - 22.5 ?= 53

45 + 8 ?= 53

53 == 53  (( YES ))

Since 53 is indeed equal to 53, then 22.5 is a solution for x.

Cheers.

8 0
2 years ago
Read 2 more answers
A cyclist has rode his bike 174 miles in 3 hours. If he continues at this rate, how long would it take him to complete another 4
In-s [12.5K]

Answer:

7 hours

Step-by-step explanation:

174/3 is 58 miles per hour so 406/58 is 7 hours

6 0
3 years ago
The value of Q is given by the relation Q = 17m. If Q and m are integers with Q greater than 150 and less than 160, then which o
zloy xaker [14]

Answer: Value of m = 9

Step-by-step explanation:

Given that the relationship between Q and m is;

Q = 17m

Make m the subject of formula

M = Q/17

If Q is greater than 150 and less than 160, then, let assume that

Q = 151, then

M = 151/17

M = 8.88

If Q = 159

M = 159/17

M= 9.35

Since m ranges from 8.88 to 9.35, the value of m = 9

6 0
2 years ago
What is the solution to the system of equations graphed below? y = –4x + 33
satela [25.4K]

y = –4x + 33

y-33 = –4x

y-33/4 = x

X=y-33/4.

5 0
2 years ago
Read 2 more answers
If perpendiculars from any point within an angle on its arms are equal, prove that it lies on the bisector of that angle
Oxana [17]
Your question can be quite confusing, but I think the gist of the question when paraphrased is: P<span>rove that the perpendiculars drawn from any point within the angle are equal if it lies on the angle bisector?

Please refer to the picture attached as a guide you through the steps of the proofs. First. construct any angle like </span>∠ABC. Next, construct an angle bisector. This is the line segment that starts from the vertex of an angle, and extends outwards such that it divides the angle into two equal parts. That would be line segment AD. Now, construct perpendicular line from the end of the angle bisector to the two other arms of the angle. This lines should form a right angle as denoted by the squares which means 90° angles. As you can see, you formed two triangles: ΔABD and ΔADC. They have congruent angles α and β as formed by the angle bisector. Then, the two right angles are also congruent. The common side AD is also congruent with respect to each of the triangles. Therefore, by Angle-Angle-Side or AAS postulate, the two triangles are congruent. That means that perpendiculars drawn from any point within the angle are equal when it lies on the angle bisector

4 0
3 years ago
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