\(\frac{1\frac{1}{5}}{3}\) | Convert the numerator to an improper fraction. |
\(\frac{\frac{5*1+1}{5}}{3}\) | |
\(\frac{\frac{6}{5}}{3}\) | Multiply by 1/3 to the numerator and denominator to eliminate this complex fraction. |
\(\frac{6}{5}*\frac{1}{3}\) | 6 and 3 have a common factor. Noticing this will make the calculations a tad easier. |
\(\frac{2}{5}\) | |
Situation 1: Robert has \(1\frac{1}{5}\) cups of fresh popcorn kernels stored in a jar, but the capacity of his popcorn machine maker is a third of the amount of popcorn kernels at a time.
In this case, \(\frac{1\frac{1}{5}}{3}\) represents the capacity, in cups, of the popcorn machine maker.
Situation 2: Robert estimates that his homework for math, science, and history will take \(1\frac{1}{5}\) hours to complete. He plans to make this homework load more manageable by working on each subject for an equal amount of time.
In this scenario. \(\frac{1\frac{1}{5}}{3}\) hours represents the amount of time he plans to work on each subject.
This is a diagram for you to reference as I solve for the radius, r.
\(C=6r+14.84\) because of what the given info states. The general formula for the relation of the radius and circumference is \(C=2\pi r\), so \(2\pi r=6r+14.84\). Now, solve for r, in feet.
\(2\pi r=6r+14.84\) | The problem states to use the approximation of \(\pi=3.14\), so substitute that into the equation. |
\(2*3.14r=6r+14.84\) | |
\(6.28r=6r+14.84\) | Subtract 6r from both sides. |
\(0.28r=14.84\) | Divide by 0.28 from both sides to finally isolate r, the variable I assigned to be the length, in feet, of the radius. |
\(r=53ft\) | |
The equation \(h=-16t^2+100t+5\) represents the height, in feet, (h) of the arrow after t seconds elapsed. The question is essentially asking after how seconds (t) will the arrow reach a height of 20 feet (h).
\(20=-16r^2+100t+5\) | Subtract 20 from both sides. |
\(-16t^2+100t-15=0\) | Use the quadratic formula to find the solutions of this equation. |
\(t=\frac{-100\pm\sqrt{100^2-4(-16)(-15)}}{2(-16)}\) | Now, simplify from here. |
\(t=\frac{-100\pm\sqrt{9040}}{-32}\) | 16 is the largest perfect square factor of 9040, so we can simplify the radical further. |
\(\sqrt{9040}=\sqrt{16*565}=\sqrt{16}\sqrt{565}=4\sqrt{565}\) | |
\(t=\frac{-100\pm4\sqrt{565}}{-32}\) | -100,4, and -32 all have a GCF of 4, so this answer can still be simplified further. |
\(t=\frac{-25\pm\sqrt{565}}{-8}\) | |
\(t_1=6.096\text{sec}\quad t_2=0.154\) | Since the question specified that the arrow had to reach the height only after reaching its maximum altitude, then the only correct answer is 6.096 seconds. |
If I am not mistaken, this problem requires you to generate the cubic equation of the four points listed. I see that \(f(0)=0\), so one of the coordinates of the x-intercept is certainly located at \((0,0)\).I also notice that there must be another zero located at \(1 because the graph crosses the x-axis there, but we cannot determine exactly where it is yet. An odd-degree polynomial cannot have an even number of real zeros (unless a multiplicity of a zero is even). We have determined already that there are 2 zeros, so a third must be lying somewhere. Yet again, we cannot determine where.
Cubic functions are written in the form \(f(x)=ax^3+bx^2+cx+d\). We can actually solve for d straightaway by plugging in \(f(0)\).
\(f(x)=ax^3+bx^2+cx+d\) | Plug in \(f(0)\) |
\(f(0)=a*0^3+b*0^2+c*0+d\) | There is a lot of simplification that can occur here! |
\(f(0)=d\) | According to your original three points, \(f(0)=0\), so substitute that in. |
\(0=d\) | |
Great! We have already solved for one of the missing variables. Our original equation is now \(f(x)=ax^3+bx^2+cx\). It only gets harder from here, though. Let's plug in the remaining points, and let's see what happens!
\(f(x)=ax^3+bx^2+cx\) | Evaluate this function at f(-1). |
\(f(-1)=a(-1)^3+b(-1)^2+c(-1)\) | Ok, now simplify. |
\(f(-1)=-a+b-c\) | By definition, \(f(-1)=15\). |
\(15=-a+b-c\) | |
Let's plug in the second point.
\(f(1)=a(1)^3+b(1)^2+c(1)\) | Simplify from here. |
\(f(1)=a+b+c\) | \(f(1)=-5\), so plug that back in! |
\(-5=a+b+c\) | |
|
Normally, I would just go right ahead and plug in the next point, but these two equations are somewhat unique.
\(\hspace{2mm}15=-a+b-c\\ -5=+a+b+c\)
Look at that! Both the a's and the c's will cancel out here! We can solve for b using elimination.
\(\hspace{2mm}15=-a+b-c\\ -5=+a+b+c\) | Add the equations together. |
\(10=2b\) | Divide by 2 on both sides! |
\(5=b\) | |
Now, let's plug in the third point.
\(f(x)=ax^3+bx^2+cx\) | Evaluate the function at \(f(2)\) |
\(f(2)=a(2)^3+b(2)^2+c(2)\) | Now, simplify. |
\(f(2)=8a+4b+2c\) | Remember that \(f(2)=12 \) and \(b=5\). |
\(12=8a+20+2c\) | Subtract 20 from both sides. |
\(-8=8a+2c\) | Divide by 2 because that is the GCF of the coefficients, and it will make the numbers nicer in the long run. |
\(-4=4a+c\) | |
Ok, let's compare the previous equation with the first one we calculated.
\(\hspace{1mm}15=-a+b-c\\ -4=4a\hspace{7mm}+c\)
It looks like more cancellation can be done, right? Well' let's do that then!
\(\hspace{1mm}15=-a+b-c\\ -4=4a\hspace{7mm}+c\) | Add both equations together. |
\(11=3a+b\) | Of course, \(b=5\). |
\(11=3a+5\) | Now, solve for a. |
\(6=3a\) | |
\(2=a\) | |
Now, plug this back into an already existing equation to find c.
\(-4=4a+c\) | \(a=2\), so substitute that value in. |
\(-4=4*2+c\) | Now, solve for c. |
\(-4=8+c\) | Subtract 8 from both sides and isolate c. |
\(-12=c\) | |
Ok, our function is now clearly defined for the points given. It is now \(f(x)=2x^3+5x^2-12x\). It is time to find the remaining zeros!
\(2x^3+5x^2-12x=0\) | Factor out a GCF first before doing anything else! | |||
\(x(2x^2+5x-12)=0\) | The next step involves finding factors of -24 that add up to 5. (-3 and 8). Break up the linear term into these portions. | |||
\(x(2x^2-3x+8x-12)=0\) | Now, factor by grouping. | |||
\(x[x(2x-3)+4(2x-3)]\) | ||||
\(x(x+4)(2x-3)=0\) | Now, set each factor equal to zero and solve. | |||
| All the zeros have been solved for finally. | |||
Therefore, the zeros are \(-4,0,1.5\).
I know why the guest got this question incorrect. Let's assume that \(m\angle ADC=90^{\circ}\), thus making it a right triangle. Let's make the assumption, too, that \(AD=DC\), thus making it an isosceles triangle.
\(\frac{\left(4\sqrt{2}\right)^2}{2}\) | This utilizing the triangle area formula. |
\(\frac{16*(\sqrt{2})^2}{2}\) | |
\(8*\sqrt{2}^2\) | |
\(8*2\) | |
\(16units^2\) | |
However, there is a key word. That word would happen to be "altitude." An altitude is a perpendicular height that extends from a vertex to the opposite side. Notice how the height of this triangle does not do this. Therefore, the original diagram does not fit the original criteria.