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3 years ago

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Mathematics

2 answers:

cupoosta [38]3 years ago

8 0

Answer:

-2 15 ...................

balu736 [363]3 years ago

6 0

5x + 93 = -6

first, you would have to subtract 93 from both sides of the equal sign

93 - 93 = 0
-6 - 93 = -99

5x = -99

next, divide by 5 on both sides

5x / 5 = x
-99 / 5 = -19.8

x = -19.8

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Divide x³-4x²+5x-2 by x-2

tatyana61 [14]

Answer:

p(x)=x^3-4x^2+5x-2

=(2)^3-4(2)^2+5(2)-2

=8-4(4)+10-2

=8-16+10-2

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4(3x^2y^4)^3 / (2x^3y^5)^4

tatyana61 [14]

Solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Step-by-step explanation:

We need to Solve the expression: $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

Solving:

$\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

$=\frac{4(3^3x^6y^{12})}{(2^4x^{12}y^{20})}\\=\frac{4(27x^6y^{12})}{(16x^{12}y^{20})}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}$

Applying exponent rule: $\frac{x^a}{x^b}=x^{a-b}$

$=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{27x^{-6}y^{-8}}{4}$

Another exponent rule says: $x^{-a}=\frac{1}{x^a}=$

$=\frac{27}{4x^{6}y^{8}}$

So, solving the expressions $\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$ we get $\frac{27}{4x^{6}y^{8}}$

Keywords: Solving Exponents

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Y-y1=m(x-x1)

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so
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Answer:

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