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3 years ago

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Why do you divide by 60 after calculating the area of the trapezoid (to calculate the distance)? I got 900 but I don’t know why

you divide by 60 to get 15 as your answer. Ignore the pen marks.

1 answer:

Free_Kalibri [48]3 years ago

6 0

Answer:

Step-by-step explanation:

What you are trying to do is a nice way to solve this problem. Without knowing any physics, you are using what math you know to get the answer. You are actually very close to being correct, and you are right. You do have to divide by 60.

Here's why and it is the one bit of physics you have to know.

The speed is km/hour. That word hour is the culprit. Your time is in minutes and you have to get it into hours. In physics, the units have to be consistent.

So ...

b1 = 19 minutes which is 19 min [1 hour / 60 minutes] = 19/60 = 0.31667 hour

b2 = 26 minutes which is 26 min [ 1 hour/60 minutes] = 26/60 = 0.4333 hour

h = 40 km/hour

Area = (0.31667 hour + 0.4333 hour)*40 km/hour /2

Area = 0.75 hour * 40 km/hr //2

Area = 15 km

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Solve for x in the equation 2x^2+3x-7=x^2+5x+39

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Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

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