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2 years ago

23. * MULTIPLE CHOICE Which is the correct factorization of -45x^2 + 20y^2

Mathematics

1 answer:

skad [1K]2 years ago

7 0

Answer:

c)-5(3x+2y)(3x-2y)

Step-by-step explanation:

1) lets find out if -45 and 20 have a GCF

-45 and 20 are both dividable by -5 so

-5(9x^2-4y^2)

1) you can see that 9 and -4 are perfect square so:

-5(3x+2y)(3x-2y)

Recheck:

1) lets use FOIL to see if this is right(if its right we will get back our original equation)

-5(9x^2-6xy+6xy-4y^2)

-5(9x^2-4y^2)

-45x^2+20y^2

hope this helps!

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A certain country's postal service currently uses 55-digit zip codes in most areas. how many zip codes are possible if there a

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There would be 100,000 if there are no restrictions on the digits and 90,000 if they cannot use 0 as the first digit.

If there are no restrictions, there are 10 possibilities for each digit:
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If the first digit cannot be 0, there are 9 possibilities for it and 10 possibilities for each of the other 4:
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2 years ago

Martha opens a new cell phone store. She spends $24,500 purchasing inventory to sell, buying furniture, and advertising her busi

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Step-by-step explanation:

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3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

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2 years ago

Which set of numbers is not closed under multiplication ?

kodGreya [7K]

Answer:

D). irrational numbers

Step-by-step explanation:

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How many handshakes are possible at a party of 65 people if everyone shakes hands once?

riadik2000 [5.3K]

You must calculate how many ways can 2 people be selected from 65.
The formula is:
n! / r! * (n-r)! =
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65*64*63! / 2 * 63! =
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Source:
http://www.1728.org/combinat.htm

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