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SSSSS [86.1K]
3 years ago
10

Which expression is equivalent to StartFraction b Superscript negative 2 Baseline Over a b Superscript negative 3 Baseline EndFr

action? Assume a not-equals 0, b not-equals 0.
Mathematics
2 answers:
pochemuha3 years ago
4 0

Answer:

\frac{b}{a}

Step-by-step explanation:

We want to find an expression that is equivalent to : \frac{b^{-2}}{ab^{-3}}

We need to use the following rule of exponents. a^{n}=\frac{1}{a^{-n}}

We apply this rule to send the b^{-3}in the denominator to the numerator to get:

\frac{b^{-2}*b^{3}}{a}

Recall also the product rule of exponents: b^m\times b^n=b^{m+n}

\implies \frac{b^{-2}*b^{3}}{a}=\frac{b^{-2+3}}{a}

We finally simplify to get:

\frac{b^1}{a}=\frac{b}{a}

PSYCHO15rus [73]3 years ago
4 0

Answer:

D

Step-by-step explanation:

I got it correct on edge

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13. The equation of line l is given as y = 0.6x – 4.
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Answer: 16.6

Step-by-step explanation:

a = slope = 0.6

b = y-intercept = -4

c = x intercept = 20/3. (Set y=0 and solve for x)

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3 0
3 years ago
Solve irrational equation pls
rusak2 [61]
\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)



\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\

4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\

There's one more condition I forgot about
-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\

Finally
x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}
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D is true!!

Hope this helped:)
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Read 2 more answers
1.Which coordinates are not of the vertices of the feasible region for the system of inequalities?
astraxan [27]
2.   I think the answer is 24

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6x + 7y <= 42
3x + 2y  <= 18    subtract:_-
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