+2446 Evaluating this expression is a matter of being especially attentive to the order of operations:
| \(-23.6+(-4.8)+(4+4/5)*0.4-(-11.5)\) | This is the original expression. Let's clean it up a bit, shall we? |
| \(-23.6-4.8+4\frac{4}{5}*0.4+11.5\) | According to order of operations, do \(4\frac{4}{5}*0.4\) first. We have to convert to an improper fraction |
| \(-23.6-4.8+\frac{24}{5}*\frac{2}{5}+11.5\) | Multiply the fractions together |
| \(-23.6-4.8+\frac{48}{25}+11.5\) | Convert \(\frac{48}{25}\)to a decimal. |
| \(-23.6-4.8+1.92+11.5\) | Do the calculations from left to right now because addition and subtraction have equal priority |
| \(-28.4+1.92+11.5\) | |
| \(-26.48+11.5\) | |
| \(-14.98\) | This is your answer |
+2446 This problem can be solved with a theorem known as secant-tangent product theorem. I'll illustrate this theorem with a table:
| Theorem in Words | Diagram | Conclusion |
| If a tangent and secant intersect in the exterior of a circle, then the product of the lengths of the secant segment and its external segments equals the length of the tangent segment squared. |  | \(BD*ED=AD^2\) |
| Secant \(\overline{BD}\) and tangent \(\overline{AD}\) intersect at point D. |
Let's apply this theorem now:
| \(YZ*YV=YX^2\) | Valid because of the secant-tangent product theorem. Plug in the values that correspond to each length and find the unknown, YX. |
| \((9+19)*9=YX^2\) | Simplify the left hand side of the equation |
| \(\sqrt{252}=\sqrt{YX^2}\) | Do the square root of both sides to get rid of the exponent and isolate YX. |
| \(|YX|=\sqrt{252}\) | Simplify the radical by finding the highest factor that is also a perfect square |
| \(|YX|=\sqrt{36*7}=6\sqrt{7}\) | Get rid of the absolute value sign by dividing your answer into the positive and negative answer |
| \(YX=\pm6\sqrt{7}\) | Of course, reject the negative answer as a negative sidelength is nonsensical in the context of geometry. This is your final answer in simplest radical form, too. |
| \(YX=6\sqrt{7}m\approx15.87m\) | I also included the decimal approximation to the hundrendth place. |
+2446 I'm pretty sure that GST stands for Goods and Services Tax. It's a Canadian tax, so the rest of the world may be unfamiliar with such:
Anyway, let's calculate the amount paid for a DVD. To find it, find how much the discount affected the price of the DVD:
| \($12.50*0.45=5.625\) | This is how much the discount affected the price. Subtract 5.625 from the original price of the DVD. |
| \(12.50-5.625=$6.875\) | In the context of money, money is always rounded to the nearest hundredth place, so round this number to that decimal. |
| \($6.88\) | |
Okay, now we know what the cost of the DVD is after discount is applied to the price. Now, let's add the discount. Normally, you would multiply 6.88 by 5% and then add it to 6.88, but there is actually a faster method that requires one fewer step: Multiply 6.88*1.05. See how that saves a step? Let's try it together:
| \($6.88*1.05=$7.224\) | Yet again, round to the nearest hundredth. |
| \($7.22\) | You are done! This is your final cost for the DVD. |
+2446 Use the law of cosines to find the missing angle measures. I'll use a picture to illustrate the law of cosines. I think is makes it easier to understand:
Source: http://hyperphysics.phy-astr.gsu.edu/hbase/imgmth/lcos.gif
Let's say that the lowercase letters have lengths. I'll arbitrarily assign them for you:
a=10
b=15
c=19
Using this information, simply substitute into the law of cosines formula. You will need a calculator that can calculate trigonometric functions:
| \(c^2=a^2+b^2-2ab\cos C\) | Simply plug in the known values and solve for the missing one. |
| \(19^2=10^2+15^2-2(10)(15)\cos C\) | Because you can use a calculator, I would not simplify anything yet. Subtract 10^2+15^2 on both sides. |
| \(19^2-10^2-15^2=-2(10)(15)\cos C\) | Divide by -2(10)(15) on both sides |
| \(\frac{19^2-10^2-15^2}{-2(10)(15)}=\cos C\) | Use the inverse cosine to isolate C. |
| \(C=\cos^{-1}(\frac{19^2-10^2-15^2}{-2(10)(15)})\) | Use your calculator to evaluate this monstrosity |
| \(C\approx96.89^\circ\) | Repeat this process for the other missing angle measures. |
One last warning before you go! Be sure that your calculator is on degree mode when doing the final calculation. Otherwise, your answer will be represented differently. In radian mode, \(\cos^{-1}(\frac{19^2-10^2-15^2}{-2(10)(15)})\approx1.69\). It is not wrong but a triangle's angle measure is usually represented in degrees--not radians. If your calculator can only preform a calculation like this in radian mode, multiply your answer by \(\frac{180}{\pi}\) to convert.
+2446 The parabola focus is the point wherein the distance to a point on a parabola is equidistant to the distance to the directrix!
To find the focus, convert the quadratic to vertex form, \(y=a(x-h)^2+k\) where \((h,k+\frac{1}{4a})\) is the focus. Let's try and do this:
| \(y=\frac{1}{8}x^2+4x+20\) | This is the original quadratic equation. In order ro convert the quadratic to the desired form above, we need to use a method called "completing the square." First, subtract 20 on both sides. |
| \(y-20=\frac{1}{8}x^2+4x\) | Multiply by 8 on both sides to get rid of the pesky fraction |
| \(8y-160=x^2+32x\) | This is where completing the square comes in handy. Do the linear x-term and half it. Take that quantity and square it. Add it to both sides. |
| \(8y-160+(\frac{32}{2})^2=x^2+32x+(\frac{32}{2})^2\) | Simplify both sides of the equation |
| \(8y+96=x^2+32x+256\) | What's the point of doing all this work? Well, the right hand side is a perfect square trinomial. |
| \(8y+96=(x+16)^2\) | Subtract 96 on both sides of the equation |
| \(8y=(x+16)^2-96\) | Divide by 8 on both sides |
| \(y=\frac{1}{8}(x+16)^2-12\) | |
Our quadratic equation is finally in vertex form. Now, we can find the focus by using the formula I mentioned above, \((h,k+\frac{1}{4a})\). Let's plug those values into this quadratic equation. First, identify what h, k, and a are.
h=-16
k=-12
a=1/8
Let's plug these values in:
| \((-16,-12+\frac{1}{4(\frac{1}{8})})\) | Do 4*1/8 first. |
| \((-16,-12+\frac{1}{\frac{1}{2}})\) | I'll use a fraction rule that states that \(\frac{a}{\frac{b}{c}}=\frac{ac}{b}\) |
| \((-16,-12+2)\) | Continue simplifying. |
| \((-16,-10)\) | |
Now, you are finally done. The point of the focus is \((-16,-10)\).
