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3 years ago

Factor out the greatest common factor from the following polynomial. 6y3 - 18xy+

Mathematics

1 answer:

melamori03 [73]3 years ago

5 0

Answer:

6y(y^2-3x)

Step-by-step explanation:

6y^3-18xy=6y(y^2-3x)

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A frightened child is restrained by her mother as the child slides down a frictionless playground slide. If the force on the chi

MissTica

Answer:

Step-by-step explanation:

Given

Restrained Force $F=116 N$

Increase in Kinetic Energy $\Delta K.E.=30 J$

Distance moved d=1.9 m

Work done by all the forces is equal to change in Kinetic Energy of Object i.e. by gravity force, Normal force and Applied Force

$\Delta K.E.=W_g+W_f+W_N$

$30=W_g-F\cdot d$

$30=W_g-116\cdot 1.9$

$W_g=250.4 J$

(b)In absence of applied force thus

work done by gravity will be equal to change in kinetic Energy

change in Kinetic Energy $=210 J$

6 0

3 years ago

You are visiting a rainforest, but unfortunately,your insect repellent has run out. As a result, at each second, a mosquito land

zmey [24]

Answer:

Step-by-step explanation:

Prob at each second, a mosquito lands on your neck =0.5

Prob it bites/it lands = 0.2

Prob it does not bite/it lands = 0.8

Hence in one second prob for getting a bite = $0.5(0.2) = 0.01$

i.e. 1/100 chance of getting bitten in one second

In 100 seconds number of bites $=1$

Expected time between successive bites = 100

6 0

3 years ago

A student uses a solution that contains 16 grams of water to conduct an evaporation experiment. At the end of one hour, the am

yanalaym [24]

Answer:

3.5% decrease means 96.5% remaining

0.965(16)

4.25% decrease means 95.75% remaining

0.9575(0.965(16))

= 14.7838 grams

5 0

3 years ago

Read 2 more answers

6j-k=23<br>3k+6j=11<br>Solve j and k using the elimination method on the simultaneous equations

frez [133]

Answer:

Step-by-step explanation:

6j - k = 23

6j + 3k = 11.....multiply by -1

---------------

6j - k = 23

-6j - 3k = -11 (result of multiplying by -1)

---------------

-4k = 12

k = -12/4

k = - 3 <=====

6j - k = 23

6j - (-3) = 23

6j + 3 = 23

6j = 23 - 3

6j = 20

j = 20/6

j = 10/3 <===

solution is : j = 10/3 and k = -3

5 0

3 years ago

For the function given state the period f(t) =6sin(3t-pi/6)-1

lapo4ka [179]

$\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}$

$\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks}\\
\quad \textit{horizontally by amplitude } |{{ A}}|\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\qquad if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\
\end{array}$

$\bf \begin{array}{llll}


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

now, with that template in mind, let's take a peek at yours

$\bf \begin{array}{lllcclll}
f(t)=&6sin(&3t&-\frac{\pi }{6})&-1\\
&\uparrow &\uparrow &\uparrow &\uparrow \\
&A&B&C&D
\end{array}\\\\
-----------------------------\\\\
period\qquad \cfrac{2\pi }{B}\iff\cfrac{2\pi }{3}$

8 0

3 years ago