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valina [46]
3 years ago
8

\times \frac{4}{7} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
torisob [31]3 years ago
5 0
Hi,
You simply need to multiply the numerator (4 in this case) with 5. Don’t touch the denominator (7)!
So the answer to this is 20 on 7 (20/7) or 2.857...
Butoxors [25]3 years ago
4 0

\bf 5\times \cfrac{4}{7}\implies \cfrac{5\cdot 4}{7}\implies \cfrac{20}{7}\implies \cfrac{14+6}{7}\implies \cfrac{14}{7}+\cfrac{6}{7}\implies 2+\cfrac{6}{7}\implies 2\frac{6}{7}

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Graph the function.<br> y = 3 square root x
s2008m [1.1K]

Answer:

x=3×3

Step-by-step explanation:

ok it's fine if you don't mind

8 0
2 years ago
Solve the equation.<br><br>-4 (y - 2) = 12<br><br>y = ?​
inna [77]
First you distribute what’s outside the parentheses (the -4)

-4•4= -4y
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New equation: -4y+8=12

Now you solve like a normal equation
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New equation: -4y=4

Now divide

Answer:-1
8 0
3 years ago
Solve: |4x + 5| + 8 &gt;(or equal to)45
KatRina [158]
This is pretty simple once you get used to this


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4x+5>_45 or 4x+5>_-45

Then solve for x for both


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8 0
3 years ago
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Virty [35]
The answer to your question is 1 7/12
8 0
3 years ago
Determine whether the improper integral converges or diverges, and find the value of each that converges.
Ksju [112]

Answer:converge at I=\frac{1}{3}

Step-by-step explanation:

Given

Improper Integral I is given as

I=\int^{\infty}_{3}\frac{1}{x^2}dx

integration of \frac{1}{x^2}  is  -\frac{1}{x}

I=\left [ -\frac{1}{x}\right ]^{\infty}_3

substituting value

I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]

I=-\left [ 0-\frac{1}{3}\right ]

I=\frac{1}{3}

so the value of integral converges at \frac{1}{3}

8 0
3 years ago
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