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We can use a graph to help with this : https://www.desmos.com/calculator/4beobd4b3x

The vertex points of the quadrilateral  are  (0, 3)  (0,2)  (1,0)  and (3, 0)

By visual inspection , the longest side  is the distance from (0,3) to (3,0)

And this distance =  √ [ (3 - 0)^2 + (3 - 0)^2 ] = √  [9 + 9 ]  = √18  =  3√2 units

V  =  (1/3)s^2h        using implicit differentiation

dV / dt   =  (2/3)s *(ds / dt) * h   +  (1/3)s^2 (dh/dt)   =  (2/3)(3)(6)*(9)  + (1/3)(3)^2 *(-1)  = 105

sin (5x - 9)  = cos (10x - 6)

Note that, because they are co-functions,    (5x - 9) + (10 x - 6)  = 90      simplify

15x - 15  =  90     add 15 to both sides

15x  = 105    divide both sides by  15

x = 7  

Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy. 

x^5y^2-x^3y+5xy^3=0

f(0)  =   a/2

To find the inverse, we have that

y =   a / [x + 2]

y  [ x + 2 ] = a

yx + 2y  =  a

yx =  a - 2y

x  =  [ a - 2y] / y       swap  x and y

y = [ a - 2x ] /x     =  the inverse = f^-1(x)

So

f^-1 (3a)  =  [ a - 2(3a) ] / [3a ]      =   [ -5a ] / 3a  =  -5/3

So  if f(0)  = f^-1(3a).....then

a/ 2   =  -5/3

a = -10/3

Are you studying Newton's Law of Cooling?    I THINK this is how to solve this Q:

T(t) = Ta + (Ti-Ta)e^-kt       Ti = initial temp    Ta = Temp ambient

T(t) = 126   =  70 + (180-70) e^-k5     results in k = .1350257

           120 = 70+ (180-70)e^-(.1350257t)         t = 5.84 minutes   =   5 minutes 50 seconds        

Here's the Asymptote code in picture form:

We can use vertical angles to fill in three of the missing angles. We have:

Since the total sum of angles in a full rotation is 360 degrees, the angles must sum up to 360. The two missing angles must be vertical angles too, so they are the same. Let us make the missing angles \(x\), and so we have:

\(2x = 360 - (2\cdot60-2\cdot40-2\cdot50)\)

\(x=50\)

Every angle is acute, so we already have 8 acute angles that we know of.

We can also combine two angles to make acute angles. The combinations below work:

30, 40

30, 50

40, 30

50, 30

Now we know 4 combinations of two angles that work.

No combinations of three angles will work because the smallest one is already obtuse.

So, the answer is 12 angles.

Yes, each has two x intercepts     Function A = 0   at x = - 2  and 1            |x| - 4  = 0   at  x = 4 and -4

the second one MIGHT be the opposite color, so you need a third sock to be certain of a match

1,  a = 32 and b = 768         (i.e., 32³ = 32,768) 

Sorry, I don't know how to work the second problem,     

and I don't understand what is being asked for in the third problem.

. 

\(\text{if }f(x)=ax^6-bx^4+x-1\text{ and }f(2)=5\text{ what is }f(-2)?\)

The answer is actually 1.

\(\begin{align*} f(2)&=a(2^6)-b(2^4)+2-1 \\ 5&=a(64)-b(16)+1 \\ 4&=a(64)-b(16) \\ \end{align*}\)

Then,

\(\begin{align*} f(-2)&=a((-2)^6)-b((-2)^4)-2-1 \\ f(-2)&=a(64)-b(16)-3~~~(\leftarrow\text{We found what }64a-16b\text{ was in the previous equation.)} \\ f(-2)&=4-3 \\ f(-2)&=1 \end{align*}\)

fixed

I think...

question 1 is -0.9

question 2 is -1/28

It is wrong. How can you factor out the 10 when m is a variable?

If Spencer mixes 11 gallons of blue paint with 11 gallons of yellow paint, then he'll have 22 gallons of green paint. 

That wasn't quite correct...

sand is better.

dirt consists of lots of organic matter which tends to trap the water into tiny pools

in good sand the water spreads homogeneously through it and maximizes the surface area of the exposed water

(I work at a horse rescue.  This winter we're all too familiar with the drainage (and lack thereof) of even mildly polluted sand)

\(f(x) = a x^6 - b x^4 + x - 1\\ \text{little trick here}\\ f_e(x) = a x^6 - b x^4 - 1 \text{ (this is the even part of }f(x))\\ f_e(x) = f_e(-x)\)

\(f(x) = x + f_e(x)\\ f(2) = 2 + f_e(2) = 5\\ f_e(2) = 3\\ f(-2) = -2 + f_e(-2) = \\ -2 + f_e(2) = \\ -2 + 3 = 1\)

\(\dfrac{A}{x+3}+\dfrac{6x}{x^2+2x-3}=\dfrac{B}{x-1}\\ \dfrac{A}{x+3}+\dfrac{6x}{(x+3)(x-1)}=\dfrac{B}{x-1}\\ \dfrac{A(x-1)+6x}{(x+3)(x-1)}=\dfrac{B(x+3)}{(x+3)(x-1)}\)

\((A+6)x-A=Bx + 3B\)

\(A+6 = B\\ -A = 3B\\ \text{Solve these two equations for }A, B\)

Never mind I forgot the product rule and yet rememberd the chain rule... for implicit too... *sigh*

Its 105

\(\text{Ok, my response didn't really apply. I misunderstood what they wanted}\\ V = 250(1.28)^d\\ V = 250(1.28)^{\frac {t}{10}}\\ V = 250\left((1.28)^{\frac{1}{10}}\right)^t\\ V = 250(1.025)^t\)

https://web2.0calc.com/questions/please-help-meee

https://web2.0calc.com/questions/please-help-meee

There is a \(\frac{2}{10}\) chance (probability of landing on a red sector) that the player wins 3 tokens, and \(\frac{2}{10} \cdot 3 = \frac{6}{10} = \frac35\).

There is a \(\frac{2}{10}\) chance (probability of landing on a green sector) that the player wins 5 tokens, and \(\frac{2}{10} \cdot 5 = \frac{10}{10} = 1\).

There is a is a \(\frac{6}{10}\) chance (probability of landing on a blue or yellow sector) that the player loses 2 tokens, and \(\frac{6}{10} \cdot (-2) = -\frac{12}{10} = -\frac{6}{5}\).

Adding these numbers up, we get \(\frac{3}{5} + 1 - \frac{6}{5} = \frac{2}{5}\), so the answer is \(\boxed{\text{C}}\).