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slava [35]
2 years ago
10

PLEASE PLEEASE HELP I PROMISE I WILL GIVE BRAINLIIEST!!!!

Mathematics
2 answers:
Sphinxa [80]2 years ago
8 0
The middle one is 76
Nadusha1986 [10]2 years ago
6 0

Answer:

76, the middle one

Step-by-step explanation:

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just sum the 2 power

6 + 2 = 8

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If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .​
Tems11 [23]

Divide through everything by <em>b</em> :

\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}

Since <em>a/b</em> < <em>c/d</em>, it follows that

\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}

Multiply through everything on the right side by <em>b/d</em> to get

\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) < <em>c/d</em>.

For the other side, you can do something similar and divide through everything by <em>d</em> :

\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}

and <em>a/b</em> < <em>c/d</em> tells us that

\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}

Then

\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) > <em>a/b</em>.

Then together we get the desired inequality.

5 0
3 years ago
Easy math not easy for me though help me help me
White raven [17]
The answer is -13
1 would become 1^2
-5 becomes -5x
and -13 stays as it is
7 0
2 years ago
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What is the completely factored form of this expression?
Eddi Din [679]

Answer:

B.

(3x + 4)(x − 7)

Step-by-step explanation:

If you multiply.

3x2-21x+4x-28

3X2-17x-28

8 0
3 years ago
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