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S_A_V [24]
3 years ago
5

Factor

y^{2}" align="absmiddle" class="latex-formula">
Mathematics
2 answers:
irinina [24]3 years ago
8 0
Sorry for the hand writing. But you want to factor out a 4y^2 which will result in (9y^2-1). Then you will factor out the equation in parentheses to (3y-1)(3y+1). Don’t forget to put the 4y^2 out front!

Nostrana [21]3 years ago
7 0

Answer:

4(3x^2-y) (3x^2+y)

Step-by-step explanation:

See image below:)

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Is 104.12 greater than 104.002
Liono4ka [1.6K]
Yes, 104.12 is greater than 104.002

because 104.12 rounds up to 104.1
        and 104.002 rounds up to 104.0


8 0
3 years ago
Which fraction is close to, but less than 1/2?
qwelly [4]

Answer:

1/9, 2/9, 3/9, 4/9, 1/8, 2/8, 3/8, 1/7, 2/7, 3/7, 1/6, 2/6, 1/5, 2/5, 1/4, 1/3.

Step-by-step explanation:

hope this helps,

8 0
3 years ago
Find the simplified product b-5/2b x b^2+3b/b-5
White raven [17]

Answer:

The product \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}

Step-by-step explanation:

Given expression \frac{b-5}{2b} and \frac{b^2+3b}{b-5}

We have to find the product of  \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}

   

Consider the given expression  \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}

Multiply fractions, we have,

\frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d}

=\frac{\left(b-5\right)\left(b^2+3b\right)}{2b\left(b-5\right)}

Cancel common factor ( b - 5 )

we have, =\frac{b^2+3b}{2b}

Apply exponent rule,

\:a^{b+c}=a^ba^c

b^2=bb

=bb+3b=b(b+3)

=\frac{b\left(b+3\right)}{2b}

Cancel common factor b , we have,

=\frac{b+3}{2}

Thus, the product  \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}

8 0
3 years ago
Read 2 more answers
Rapid Rental Car charges a $45 rental fee, $15 for gas, and $0.75 per mile driven. For the same car, Capital
olga nikolaevna [1]

Your total cost will be 80

3 0
2 years ago
Read 2 more answers
1/2a=1/4a+2<br> Solve for a:<br> Check your solution:
vekshin1

Answer:

a=-1

Step-by-step explanation:

1/2a=1/4a+2

by cross multiplying,

2a=4a+2

2a-4a=2

-2a=2

a=2/-2

a=-1

hope you understood  :)

7 0
3 years ago
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