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Effectus [21]
3 years ago
9

Please ignore my work. Can you please solve and show all work, thanks

Mathematics
1 answer:
valina [46]3 years ago
3 0

Answer:

Answer is (1): 2 times (x^2 -13)

Step-by-step explanation:

Using the FOIL method, you can conclude that (x-5)(x+5) is X^2 -5x +5x -25. Since the two middle terms cancel, you are left with X^2 -25. For the right side of the equation, (x-1)(x+1), using the FOIL method gets you x^2 -x +x -1. Once again, the -x and x cancel out and you end up with X^2 -1. To add both of these together, you can get rid of all parentheses and do the work. Two x^2 added together is 2X^2, and -25-1 is negative 26. After this, you end up with 2x^2 -26. To get to your final answer you need to take out the GCF of both terms, (the largest number that can go into 2x^2 and -26), which is 2. Taking the two out leaves us with 2(x^2 -13), which is your final answer. It looks like you were on the right track.

FOIL Property

Multiply first terms

Then Outside terms

Then inside terms

Then the last terms

Hope this helps!

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Can someone help me round these two numbers? I completely forgot how ‍
tigry1 [53]

Answer:

Step-by-step explanation:

If you are rounding to nearest tenth

2.6  1.9

Nearest hundredth

2.6   1.87

If the number to the right of the one you want to round is 5 or above, you go up a number. If it's below, you go down 1.

Hope this helps.

5 0
3 years ago
What is the solution for the equation StartFraction 5 Over 3 b cubed minus 2 b squared minus 5 EndFraction = StartFraction 2 Ove
wolverine [178]

Answer:

The solutions are:

b=0,\:b=4

Step-by-step explanation:

Considering the expression

  • \frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

Solving the expression

\frac{5}{3b^3-2b^2-5}=\frac{2}{b^3-2}

\mathrm{Apply\:fraction\:cross\:multiply:\:if\:}\frac{a}{b}=\frac{c}{d}\mathrm{\:then\:}a\cdot \:d=b\cdot \:c

5\left(b^3-2\right)=\left(3b^3-2b^2-5\right)\cdot \:2

5b^3-10=6b^3-4b^2-10

\mathrm{Switch\:sides}

6b^3-4b^2-10=5b^3-10

6b^3-4b^2-10+10=5b^3-10+10

6b^3-4b^2=5b^3

\mathrm{Subtract\:}5b^3\mathrm{\:from\:both\:sides}

6b^3-4b^2-5b^3=5b^3-5b^3

b^3-4b^2=0

Using\:the\:Zero\:Factor\:Principle: if\:\mathrm ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

So,

b=0,b-4=0

b=0,b=4

Therefore, the solutions are:

b=0,\:b=4

4 0
3 years ago
Read 2 more answers
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3 years ago
If 1/2 of 5 is 3, then what is 1/3 of 10?
Dahasolnce [82]
3.3 repeated hope this helps
3 0
3 years ago
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Focus on the regression line. Notice how (0,90) and (1,75) are on this line. Compute the slope of the line through those points

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