All categories

2 years ago

helpppp plsss What shape is created by the rotation around the given axis "a"? Prism Cone Pyramid Sphere

Mathematics

1 answer:

coldgirl [10]2 years ago

3 0

Answer:

Option (2)

Step-by-step explanation:

From the picture attached,

When the given shape is rotated about the axis 'a',

Triangle ABC forms a 3-D figure of a cone.

Segment AB is the diameter of the circular base and CD is the height of the cone.

Therefore, Option (2) CONE will be the answer.

You might be interested in

Help me. please <3....

AVprozaik [17]

2
3
-7
24
-25
-3
second row
-12
-2
-9
-38
10
-28

5 0

2 years ago

Read 2 more answers

Katie keeps a record of her transactions in a notebook. Looking over her records for the past two weeks, Katie finds that her ba

kodGreya [7K]

<span>c is the correct answer</span><span />

4 0

3 years ago

Read 2 more answers

22= 18x + 4 please help!!

vredina [299]

Answer:

x=1

Step-by-step explanation:

22-18=4 18-18=/

18x1=18

22=18(1)+4

22-18(1)=4

7 0

2 years ago

Read 2 more answers

Noah and Lin are each trying to solve the equation x² - 6x + 10 = 0. They know that the solutions to x² = - 1 are i and -i, but

Leviafan [203]

Answer:

Solving the equation $x^2 - 6x + 10 = 0$ we get $x=3+i \ , \ x=3-i$

Step-by-step explanation:

Noah's step to solve the equation $x^2 - 6x + 10 = 0$ are:

$x^2 - 6x + 10 = 0\\x^2 - 6x = -10\\x^2 - 6x + 9 = -10 + 9\\(x-3)^2 = -1$

The next step is to take square root on both sides because: $\sqrt{x^2}=x$

$\sqrt{(x-3)^2}=\sqrt{-1} \\x-3=\pm\sqrt{-1} \\x=\pm(\sqrt{-1})+3 \\$

Now, we know that: $\sqrt{-1}=i$

$x=\pm(i)+3\\x=3+i \ , \ x=3-i$

So, Solving the equation $x^2 - 6x + 10 = 0$ we get $x=3+i \ , \ x=3-i$

8 0

3 years ago

The question is below answer please

liraira [26]

1. 51
2. 10
3. 14
4. 7
5. 13
6. 6
7. 4
8. 22
9. 29
10. 36
You just have to cross multiply and then divide it by 100

7 0

3 years ago