Square root of 48a^6b^5

Mathematics

1 answer:

DiKsa [7]1 year ago

3 0

Answer:

4a³b²√(3b)

Step-by-step explanation:

Even powers of any of the factors can be brought out from under the square root radical.

$\sqrt{48a^6b^5}=\sqrt{(4^2a^6b^4)(3b)}=\sqrt{(4a^3b^2)^2(3b)}=\boxed{4a^3b^2\sqrt{3b}}$

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How to estimate to the nearest tens 671×32

aleksandrvk [35]

20, 100

Step-by-step explanation:

You look at the tens column, in this case the numbers would be "7" in the "671" and the "3" in the "32". If you're rounding the tens column you look at the ones column and if the number is bellow 5 you round down, if its 5 nd over you round up. Since the number is 1 in the ones column that means the 7 will round down so it would be "670", for "32" the number is "2" (which is bellow 5) so we will round down again to "30". This will leave you with the equation 670×30 which is 20, 100.

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2 years ago

What sine function represents an amplitude of 1, a period of 2π, a horizontal shift of π, and a vertical shift of −4?

Gwar [14]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\
% function transformations for trigonometric functions
\begin{array}{rllll}
% left side templates
f(x)=&{{ A}}sin({{ B}}x+{{ C}})+{{ D}}
\\\\
f(x)=&{{ A}}cos({{ B}}x+{{ C}})+{{ D}}\\\\
f(x)=&{{ A}}tan({{ B}}x+{{ C}})+{{ D}}
\end{array}
\\\\
-------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{function period or frequency}\\
\left. \qquad \right. \frac{2\pi }{{{ B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\
\left. \qquad \right. \frac{\pi }{{{ B}}}\ for\ tan(\theta),\ cot(\theta)$

with that template in mind, let's see

the period of 2π, well, for that, you have to do nothing, because that's sine original period

Amplitude of 1, well, that's also sine's original amplitude

horizontal shift/phase... ok, that means C/B = π, now, it doesn't say if to the left or the right, so I gather either is ok, so...let's do the right

you could use then, C = -π and B = 1, that gives you -π/1 or -π

vertical shift of -4, well, that simply means D = -4

f(θ) = sin(θ - π) - 4 <-- will do

5 0

3 years ago

Solve for a, if F= w/9a <br> need step by step explanation

Law Incorporation [45]

Answer:

a=w/9F

Step-by-step explanation:

First multiply 9a to F to have 2 equations equal each other. Divide 9aF by 9F to get a on it’s own.

Steps clear down below