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The absolute value of a number is the distance it is from 0 onthe number line. 

Distances are never negative. so they can never be less than -5.

So there are no solutions to this .... not in the real number system anyway.

A sine function has the following key features:

Period = π

Amplitude = 2

Midline: y=−2

y-intercept: (0, -2)

The function is a reflection of its parent function over the x-axis.

Use the sine tool to graph the function. The first point must be on the midline and the second point must be a maximum or minimum value on the graph closest to the first point.

the parent funtion is     y=sinx

this has a period of 2pi, and amplitude of 1 and a midline of y=0

\(y=asin[n(x+c)]+m\\ \text{a is the amplitude}\\ \text{m is the midline.}\\ \text{c is the phase shift that you probably don't need to worry about, it is zero}\\ \text{the period is }\frac{2\pi}{n} \)

Since it is reflected over the x axis, the a value will be negative so

Period = π                           n=2

Amplitude = 2                     a= - 2    (becasue of the xaxis reflection)

Midline: y=−2                      m=-2

y-intercept: (0, -2)               hopfully this one will just work.

Can you use that formula I just gave you to work out the equation of your sine curve?

Then graph it in desmos and check that it works :)

\((3x+3)(x-4)=3(x+1)\\ 3x^2-12x+3x-12=3x+3\\ 3x^2-9x-12=3x+3\\ 3x^2-9x-12-3x-3=3x+3-3x-3\\ 3x^2-12x-15=0\\ \text{divide both sides by 3}\\ x^2-4x-5=0\\ (x-5)(x+1)=0\)

now if 2 things multiply together to give 0 then one of them, or both of them, must equal 0.

so

x-5=0      or      x+1=0

x=5         or         x=-1

I do not understand what the question is saying..... 

The bouy is at the surface isn't it? Why do we care about the ocean depth?   I don't get it.

measures of central tendency are   mean, mode and median.

Start by putting all thes scores from smallest to biggest.  

the mean is just the average, add them all up and then divide by how many there are.

The median is the one or average of the two that are in the middle

the mode is the score that haapens the most.  (there can be any number of modes)

measures of dispersion?

range, interquatile range, standard deviation.   Ar they the ones that you want?  Do you know how to find them?

(3x+3)(x-4)=3(x+1)

 If $\bold{a}$ and $\bold{b}$ are vectors such that $\|\bold{a}\| = 4$, $\|\bold{b}\| = 5$, and
 $\|\bold{a} + \bold{b}\| = 7$, then find $\|2a-3b\|$

If a and b are vectors such that \(||a||=4\), \(||b||=5\), and \(||a+b||=7\), then find \(||2a-3b||\).

1.) trigonometric
\(x = ||2\vec{a}-3\vec{b}||\)

\(\begin{array}{|rcll|} \hline 7^2 &=& 4^2+5^2- 2*4*5*\cos(A) \\ 49 &=& 16+25- 40\cos(A) \\ 40\cos(A) &=& 16+25-49 \\ 40\cos(A) &=& -8 \\ \mathbf{\cos(A)} &\mathbf{=}& \mathbf{-\dfrac{1}{5}} \\\\ x^2 &=&(2\cdot 4)^2+(3\cdot 5)^2-2*8*15*\cos(180^{\circ}-A )\\ x^2& =&8^2+15^2-16*15*\cos(180^{\circ}-A )\\ x^2&=& 64+225 + 240*\cos(A) \\ x^2&=& 289 + 240*\left(-\dfrac{1}{5}\right) \\ x^2&=& 289 - 48 \\ x^2&=& 241 \\ \mathbf{x }&\mathbf{=}& \mathbf{\sqrt{241 }} \\ \hline \end{array} \)

2.) vectorial

\(\text{Let $\vec{a}=\dbinom{x_a}{y_a} $ } \\ \text{Let $\vec{b}=\dbinom{x_b}{y_b} $ } \\ \text{Let $\vec{a}\cdot \vec{b}= x_a\cdot x_b+y_a\cdot y_b $ } \\ \text{Let $\vec{a}+\vec{b}=\dbinom{x_a+x_b}{y_a+y_b} $ } \\ \text{Let $2\vec{a}-3\vec{b}=\dbinom{2x_a-3x_b}{2y_a-3y_b} $ } \)

\(\begin{array}{|rcll|} \hline ||\vec{a}+\vec{b}|| &=& 7 \\ ||\vec{a}+\vec{b}||^2 &=& 7^2 \\ ||\vec{a}+\vec{b}||^2 &=& 49 \quad | \quad ||a+b||^2 = (x_a+x_b)^2 + (y_a+y_b)^2 \\ (x_a+x_b)^2 + (y_a+y_b)^2 &=& 49 \\ x_a^2 +2x_ax_b+x_b^2 + y_a^2 + 2y_ay_b+y_b^2 &=& 49 \\ x_a^2+y_a^2+x_b^2+y_b^2 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_a^2+y_a^2 = ||\vec{a}||^2=4^2=16 \\ 16+x_b^2+y_b^2 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_b^2+y_b^2 = ||\vec{b}||^2=5^2=25 \\ 16+25 +2(x_ax_b+ 2y_ay_b) &=& 49 \quad | \quad x_a\cdot x_b+y_a\cdot y_b = \vec{a}\cdot \vec{b}\\ 41 +2\vec{a}\cdot \vec{b} &=& 49 \\ 2\vec{a}\cdot \vec{b} &=& 8 \\ \mathbf{\vec{a}\cdot \vec{b}} &\mathbf{=}& \mathbf{4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline ||2\vec{a}-3\vec{b}|| &=& ||\dbinom{2x_a-3x_b}{2y_a-3y_b} || \\ ||2\vec{a}-3\vec{b}||^2 &=& (2x_a-3x_b)^2 + (2y_a-3y_b)^2 \\ &=& 4x_a^2 -12x_ax_b + 9 x_b^2 + 4y_a^2 - 12y_ay_b + 9 y_b^2 \\ &=& 4(x_a^2+ y_a^2) + 9( x_b^2+y_b^2) -12(x_ax_b + y_ay_b) \\\\ && x_a^2+y_a^2 = ||\vec{a}||^2=4^2=16 \\ && x_b^2+y_b^2 = ||\vec{b}||^2=5^2=25 \\ && x_a\cdot x_b+y_a\cdot y_b = \vec{a}\cdot \vec{b}\\ \\ &=& 4\cdot 16 + 9\cdot 25 -12\vec{a}\cdot \vec{b} \quad | \quad \vec{a}\cdot \vec{b} =4\\ &=& 64 + 225 -12\cdot 4 \\ &=& 289 - 48\\ &=& 241 \\ \mathbf{||2\vec{a}-3\vec{b}||} &\mathbf{=}& \mathbf{\sqrt{241 }} \\ \hline \end{array}\)

She got back to me!

So the three angles beneath the sliding cable all equal 60 degrees, equalling 180 degrees total.
Hope that helps! I can further clarify if needed

3/5 < 12 / x         cross-multiply

3x< 5*12

3x < 60              divide both sides by 3

x < 20

So.... A. 19

What's the calculation?

whoops, i just realized :P

I could try, but you need to define p,q,r,s.

are you trying to go for their coordinates?

in that case, I can try barycentric coordinates.

7w+17-12w=   -38           subtract 17 from both sides

7w - 12w  = -55      

-5w = - 55       divide both sides by -5

w = 11

\(\text{using }n \text{ subintervals we get}\\ x_k = \dfrac{(3-0)k}{n} = \dfrac{3k}{n}\\ \Delta = x_1- x_0 = \dfrac 3 n\)

\(\text{the left hand Riemann sum is} \\ \dfrac 3 n\sum \limits_{k=0}^{n-1}~\dfrac{3k}{n} + \dfrac{9k^2}{2n^2} = \dfrac{9}{4 n^2}-\dfrac{45}{4 n}+9\\ \text{you need to show this last equality of course}\)

\(\text{Pretty clearly}\\ \lim \limits_{n\to \infty}~\dfrac{9}{4n^2}-\dfrac{45}{4n} + 9 = 9\\ \text{which is the value of the definite integral of }y \text{ over }[0,3]\)

Eh.....as EP said......many on here consider parentheses/brackets to be unnecessary "frills"

\(\text{if }k \neq 0 \text{ then}\\ g^{-1}(x) = \dfrac 1 k f^{-1}(x)\\ \text{so the only value of }k \text{ for which }g(x) \text{ is not invertible is }k=0\\ \text{the sum of }0 \text{ is }0\)

\(\begin{pmatrix}x\\y\end{pmatrix} = \begin{pmatrix}2\\-3\end{pmatrix} t + \begin{pmatrix}1\\2\end{pmatrix}\)

\(\text{thus we are looking for a scaled version of the vector }\begin{pmatrix}2\\-3\end{pmatrix}\)

\( \begin{pmatrix}1&1\end{pmatrix}\cdot c\begin{pmatrix}2\\-3\end{pmatrix}=3\\ -c=3\\ c=-3\\ \begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}-6\\9\end{pmatrix}\)

I always forget about L'hopital's rule - thanks for you answer and thanks for reminding me.

Bur....

Why not just use l'hopital's rule to start with?

\(\displaystyle\lim_{x\rightarrow 1}\;\;\frac{xlnx}{1-x}\\ =\displaystyle\lim_{x\rightarrow 1}\;\;\frac{lnx+1}{-1}\\ =-1\)

If you know the current draw of the motor , it might be simpler to in-line a resistor with the motor and skip the IC... (like a 20 ohm resistor if your motor draws 200 mAmp)

FYI from the  7805 spec page   ( you might want to read it over)

https://electronicsforu.com/resources/learn-electronics/7805-ic-voltage-regulator

The greater the difference between the input and output voltage, more the heat generated. If the regulator does not have a heat sink to dissipate this heat, it can get destroyed and malfunction. Hence, it is advisable to limit the voltage to a maximum of 2-3 volts above the output voltage. So, we now have 2 options. Either design your circuit so that the input voltage going into the regulator is limited to 2-3 volts above the output regulated voltage or place an appropriate heatsink, that can efficiently dissipate heat.

I'm Dandy...sooo happy to see that you decided to still use this website after all. I mean despite you saying its trash and all I'm so glad to see that you've decided to give it another try aha!

See:

https://web2.0calc.com/questions/pls-help-fast_1

+ 9v     goes to the IC (input)  and the   cooler +

-  9v    terminal gos to the 'ground' of the IC and the cooler -

'output' of the IC goes to the fan......then the other end of the fan gets connected to the ground (-9v wire) also.....

    I do not have access to a scanner right now.....hope you can draw it....

Hope your battery has enough 'oomph' to run the cooler and fan and IC.    The IC has a max output of 5v 1 amp......hope your fan requires less than 1 amp.

physics does have a lot of math involved in it, right?