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Vaselesa [24]
2 years ago
9

Helppppppppppp plzzzz

Mathematics
2 answers:
Dovator [93]2 years ago
7 0

3.14*r^2

Hope this helps!
Fed [463]2 years ago
3 0
3.14*r^2

Hope it helps you solve them
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Point E is located at (2, −3), and point F is located at (−2, −1). Find the point that is 3 over 4 the distance from point E to
irakobra [83]
It woud be very far.<span>Explain how you would graph the line containing a slope of –1/5 that goes through the point (1,–4).</span>
3 0
3 years ago
Bob and mark talk about their families.bob says he has three kids, the product of Their age is 72. He gives another clue: the su
drek231 [11]
<span>Bob says he has 3 kids, the product of their age is 72.
So the possible ages of Bob's kids can be taken from the factors of 72

72 = 2 * 2 * 2 * 3 * 3

</span><span>Finally, bob says, my youngest child called justice.
</span>So meaning, Bob has a youngest child. He didn't mention about a twin.

So the possible ages the following:
1, 2, 36 --> sum: 39
1, 4, 18 --> sum: 23
1, 8, 9 --> sum: 18

<span>Once the sum of the ages will be revealed, then the possible ages are mentioned above.</span>
8 0
2 years ago
Which equation can be used to determine the distance between the origin and (-2,-4)?
loris [4]

Answer:

I believe it is the second 1

5 0
2 years ago
A tree casts a shadow of 3 meters, while a meter stick casts a shadow of 0.75 meters. How tall is the tree?
Lunna [17]
3.75 meters I would presume.

Explanation: since a single meter casts a shadow of 0.75 meters that means for each meter, 0.25 meters are missing. So I added 0.75 to three, because there was 3 meters and 3 time 0.25 was 0.75 .
6 0
3 years ago
A box contains 8 two-inch screws. Four have a Phillips head and 4 have a slotted head. In how many ways can 4 screws be chosen s
hammer [34]

Answer:

There are 6\cdot 6=36 different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

Step-by-step explanation:

If 4 screws must be chosen so that 2 have a Phillips head and 2 have a slotted head, then you have to choose 2 screws with a Phillips head from 4 screws with a Phillips head and 2 screws with a slotted head from 4 screws with a slotted head.

You can choose 2 screws with a Phillips head from 4 screws with a Phillips head in

C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6

different ways.

You can choose 2 screws with a slotted head from 4 screws with a slotted head in

C^4_2=\dfrac{4!}{2!(4-2)!}=\dfrac{4!}{2!\cdot2!}=\dfrac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot1\cdot2}=6

different ways.

In total there are 6\cdot 6=36 different ways to choose 4 screws such that 2 have a Phillips head and 2 have a slotted head.

7 0
2 years ago
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