Please helppp number 11

Mathematics

1 answer:

aivan3 [116]2 years ago

4 0

Answer:

(1/3)⁷

Step-by-step explanation:

negative exponents work like this:

x⁻²=(1/x)²

Apply this theory to this equation:

3⁻⁷=

(1/3)⁷

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Can you help out on this please

kari74 [83]

Answer:

Step-by-step explanation:

Given

9² × ( $\frac{1}{3}$ )² - 9

= 81 × $\frac{1}{9}$ - 9 ← cancel 81 and 9 by 9 leaving

= 9 - 9

= 0 → B

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Read 2 more answers

Solve the equation:

PilotLPTM [1.2K]

Answer:

Step-by-step explanation:

$4\sqrt{3-\frac{1}{x} } -\sqrt{\frac{x}{3x-1} } =3\\4\sqrt{\frac{3x-1}{x} } -\sqrt{\frac{x}{3x-1} } =3\\\\put ~\sqrt{\frac{3x-1}{x}} =a\\4a-\frac{1}{a} =3\\4a^2-1-3a=0\\4a^2-4a+1a-1=0\\4a(a-1)+1(a-1)=0\\(a-1)(4a+1)=0\\a=1,-\frac{1}{4}\\$

$\sqrt{\frac{3x-1}{x} } =1\\3x-1=x\\3x-x=1\\2x=1\\x=\frac{1}{2}$

when a=-1/4

$\sqrt{\frac{3x-1}{x} } =-\frac{1}{4} \\$

$\frac{3x-1}{x} =\frac{1}{16} \\48x-16=x\\47x=16\\x=\frac{16}{47}$

$\frac{(1-2x)^2 \times (x^2-9)}{-x^2-x+6} \geq 0,\\both ~numerator ~and~denominator ~are~of~same~sign.\\\frac{(1-2x)^2(x^2-9)}{-x^2-3x+2x+6} \geq 0\\\frac{(1-2x)^2(x+3)(x-3)}{-x(x+3)+2(x+3)} \geq 0\\\frac{(1-2x)^2(x+3)(x-3)}{(x+3)(-x+2)} \geq 0\\(1-2x)^2\geq 0~(always)\\x\neq -3\\case.~1.\\both~numerator~and~denominator >0\\\frac{x-3}{-x+2} \geq 0\\x-3\geq 0\\x\geq 3\\-x+2> 0\\-x>-2\\x$