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viva [34]
4 years ago
7

A part of a line with endpoints on both ends is a(n):

Mathematics
1 answer:
patriot [66]4 years ago
4 0

I believe it's a segment.

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Is 7,434 divisible by 6 true are false
kogti [31]
<span>7,434 is divisible by 6 so this is true

</span>
3 0
3 years ago
One number is 6 more than another. The difference between their squares is 204. What are the numbers?
Crazy boy [7]

Answer:

14 and 20

Step-by-step explanation:

<u><em> Let x and y be two numbers where x < y</em></u><em>.</em>

We are given :

y - x = 6

y² - x² = 204

Then

y - x = 6

(y - x)(y + x) = 204

Then

y - x = 6

6 × (y + x) = 204

then

y - x = 6       (Equation 1)

y + x = 34    (Equation 2)

Then

(Equation 1) + (Equation 2) ⇒ 2y = 40 ⇒ y = 20

Now ,we substitute y by 20 in equation 1 :

y - x = 6

⇔ 20 - x = 6

⇔ 20 - 6 = x

⇔ x = 14

4 0
1 year ago
Read 2 more answers
1. Simplify: (6-√5)(5+√5)
aniked [119]
150 because 6 times 5 is 30 then you multiply whats inside the square root thing so you'd get 25, 25 is a perfect square which simplifies to 5 then 5 times 30 is 150.
7 0
3 years ago
Read 2 more answers
What is x? Super confused
Over [174]

Answer:

\\ from \: triangle \: applying \: pyth \\ \sqrt{ {3}^{2}  +  {3}^{2} }  = 3 \sqrt{2}  \\ from \: rectangle \:applying \: pyth \\ {x}^{2} = {7}^{2} - 3 \sqrt{2}  \: units

6 0
3 years ago
Prove the following statement.
gayaneshka [121]

Answer:

You can prove this statement as follows:

Step-by-step explanation:

An odd integer is a number of the form 2k+1 where k\in \mathbb{Z}. Consider the following cases.

Case 1. If k is even we have: (2k+1)^{2}=(2(2s)+1)^{2}=(4s+1)^{2}=16s^2+8s+1=8(2s^2+s)+1.

If we denote by m=2s^2+2 we have that (2k+1)^{2}=8m+1.

Case 2. if k is odd we have: (2k+1)^{2}=(2(2s+1)+1)^{2}=(4s+3)^{2}=16s^2+24s+9=16s^{2}+24s+8+1=8(2s^{2}+3s+1)+1.

If we denote by m=2s^{2}+24s+1 we have that (2k+1)^{2}=8m+1

This result says that the remainder when we divide the square of any odd integer by 8 is 1.

6 0
3 years ago
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