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3 years ago

2x10^-4 - 1x10^-5 show work. I got an answer of 1.9x10^-4,

2 answers:

8 0

$\bf \left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{ n}}}\implies \cfrac{1}{a^{-{{ n}}}}\\\\
-------------------------------$

$\bf 2\times 10^{-4}-1\times 10^{-5}\implies 2\cdot \cfrac{1}{10^4}-1\cdot \cfrac{1}{10^5}\impliedby 
\begin{array}{llll}
\textit{using the LCD}\\
of~10^5
\end{array}
\\\\\\
\cfrac{2}{10^4}-\cfrac{1}{10^5}\implies \cfrac{(10\cdot 2)~-~(1\cdot 1)}{10^5}\implies \cfrac{20-1}{10^5}\implies \cfrac{19}{10^5}
\\\\\\
19\cdot \cfrac{1}{10^5}\implies 19\times 10^{-5}$

STALIN [3.7K]3 years ago

5 0

2 • 10^-4 – 1 • 10^-5 =
0.0002 – 0.00001 = 0.00019
1.9 • 10^-4

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Factor -1/4 out of -1/2x -5/4y. (i will mark brainlyest)

mash [69]

Answer:

-1/4 (2x - 5y)

Step-by-step explanation:

For us to factor -1/4 out of -1/2x -5/4y

We will have to multiply the values in bracket by the reciprocal of -1/4 i.e -4 as shown

-1/2x - 5/4 y

= -1/4 (-4(-1/2x) - (-4)(-5/4y))

= -1/4 (4/2 x - 20/4 y)

= -1/4 (2x - 5y)

Hence the factored expression is -1/4 (2x - 5y)

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Which equation has the solutions x=1+or-square root of 5?

stiv31 [10]

We will proceed to solve each case to determine the solution of the problem.

case a) $x^{2}+2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=-4+1$

$x^{2}+2x+1=-3$

Rewrite as perfect squares

$(x+1)^{2}=-3$

$(x+1)=(+/-)\sqrt{-3}\\(x+1)=(+/-)\sqrt{3}i\\x=-1(+/-)\sqrt{3}i$

therefore

case a) is not the solution of the problem

case b) $x^{2}-2x+4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=-4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=-4+1$

$x^{2}-2x+1=-3$

Rewrite as perfect squares

$(x-1)^{2}=-3$

$(x-1)=(+/-)\sqrt{-3}\\(x-1)=(+/-)\sqrt{3}i\\x=1(+/-)\sqrt{3}i$

therefore

case b) is not the solution of the problem

case c) $x^{2}+2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}+2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}+2x+1=4+1$

$x^{2}+2x+1=5$

Rewrite as perfect squares

$(x+1)^{2}=5$

$(x+1)=(+/-)\sqrt{5}\\x=-1(+/-)\sqrt{5}$

therefore

case c) is not the solution of the problem

case d) $x^{2}-2x-4=0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$x^{2}-2x=4$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$x^{2}-2x+1=4+1$

$x^{2}-2x+1=5$

Rewrite as perfect squares

$(x-1)^{2}=5$

$(x-1)=(+/-)\sqrt{5}\\x=1(+/-)\sqrt{5}$

therefore

case d) is the solution of the problem

therefore

the answer is

$x^{2}-2x-4=0$

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