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3 years ago

Perform the indicated operation. (-18)/-(4)

1 answer:

7 0

A negative divided by a negative is always a positive. 4 can go into 18 only 4 times to get 16 with a remainder of 2. 2 is half of four, therefore -4 can go into -18 4.5 times.
The answer is 4 1/2 or 4.5.
I hope this helps ^-^

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Can anyone help me with inverse function please?

12345 [234]

${h}^{ - 1} oh(x) = x \\ {h}^{ - 1} (4x + m) = x \\ 2k(4x + m) + \frac{5}{8} = x \\ 8kx + 2km + \frac{5}{8} = x$

k and m must satisfy 8k = 1 mean k = ⅛

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3 years ago

Can someone please explain?

geniusboy [140]

This is an isosceles triangle. The angles at the base are congruent (they have the same measures).

Look at the picture.

We know: The sum of the measures of angles in a triangle is equal 180°.

Therefore we have the equation:

$2x-10+2x-10+3x+25=180$ <em>combine like terms</em>

$(2x+2x+3x)+(-10-10+25)=180$

$7x+5=180$ <em>subtract 5 from both sides</em>

$7x=175$ <em>divide both sides by 7</em>

$x=25$

$m\angle A=2x-10\to m\angle A=2(25)-10=50-10=40\\\\m\angle C=m\angle A=40\\\\m\angle B=3x+25\to m\angle B=3(25)+25=75+25=100^o$

<h3>Answer: D. m∠C = 40</h3>

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3 years ago

Help me? idk the answer :P

Alexus [3.1K]

$\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}}}}\\\\
-------------------------------\\\\
\left( 2^8\cdot 3^{-5}\cdot 6^0 \right)^{-2}\left( \cfrac{3^{-2}}{2^3} \right)^4\cdot 2^{28}\impliedby \textit{let's do the first group}
\\\\
-------------------------------\\\\
$

$\bf \left( 2^8\cdot \cfrac{1}{3^5}\cdot 1 \right)^{-2}\implies \left( \cfrac{2^8}{3^5} \right)^{-2}\implies \left( \cfrac{3^5}{2^8} \right)^{2}\implies \cfrac{3^{2\cdot 5}}{2^{2\cdot 8}}\implies \boxed{\cfrac{3^{10}}{2^{16}}}\\\\
-------------------------------\\\\
\textit{now the second group}\qquad \left( \cfrac{3^{-2}}{2^3} \right)^4\implies \left( \cfrac{\frac{1}{3^2}}{2^3} \right)^4\implies \left( \cfrac{1}{2^3\cdot 3^2} \right)^4$

$\bf \cfrac{1^4}{2^{4\cdot 3}\cdot 3^{4\cdot 2}}\implies \boxed{\cfrac{1}{2^{12}\cdot 3^8}}\\\\
-------------------------------\\\\
\textit{so we end up with\qquad }\cfrac{3^{10}}{2^{16}}\cdot \cfrac{1}{2^{12}\cdot 3^8}\cdot 2^{28}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{16}\cdot 2^{12}\cdot 3^8}
\\\\\\
\cfrac{3^{10}\cdot 2^{28}}{2^{16+12}\cdot 3^8}\implies \cfrac{3^{10}\cdot 2^{28}}{2^{28}\cdot 3^8}\implies 3^{10}\cdot 2^{28}\cdot 2^{-28}\cdot 3^{-8}$

$\bf 3^{10-8}\cdot 2^{28-28}\implies 3^2\cdot 1\implies 3^2\implies 9$

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3 years ago