Posts tagged with: OT Test 6 Section 4

Practice Test 6 Calculator Section # 29

All you need to know about average speed is that it’s equal to \dfrac{\text{distance traveled}}{\text{time it took to travel that distance}}.

You’re given a formula for the distance traveled: when the car’s been traveling t seconds, it’s gone a distance of 16t\sqrt{t}. That formula contains all you need to calculate the average speed.

\text{distance traveled}=16t\sqrt{t}

\text{time it took to travel that distance}=t

\text{average speed}&=\dfrac{\text{distance traveled}}{\text{time it took to travel that distance}}=\dfrac{16t\sqrt{t}}{t}=16\sqrt{t}

Test 6 Section 4 #32

This question is built to nail you if you forget to distribute. The first step is the most important.

    \begin{align*}2(5x-20)-(15+8x)&=7\\10x-40-15-8x&=7\end{align*}

You need to remember to both distribute the 2 across the 5x and the –20 and also distribute the negative across the 15 and the 8x. Once you’ve done that, the rest is just plain-old vanilla solving for x. Here are my steps:

    \begin{align*}2(5x-20)-(15+8x)&=7\\10x-40-15-8x&=7\\2x-55&=7\\2x&=62\\x&=31\end{align*}

Could you solve Test 6, Section 4, number 18 for me? Thanks!

Sure thing. You’ve got your riser-tread formula: 2hd = 25. Now you’re told that d must be at least 9 and h must be at least 5. Translating that into math: d ≥ 9 and h ≥ 5.

Of course, since 2hd must always equal 25, there are also upper bounds on the values of d and h—they can’t be too small, but they can’t be too big, either. In fact, as one gets bigger, it forces the other one to get smaller.

Let’s plug in 9 for d (that’s its minimum value) and see what we’d get for h:

2hd = 25
2h + 9 = 25
2h = 16
h = 8

What would happen if we plugged 10 in for d instead?

2h + 10 = 25
2h = 15
h = 7.5

Ahh, see? If d gets bigger, it makes h smaller. So the BIGGEST value we can have for h is 8 if we have a minimum value for d at 9.

We now know that h must be at least 5, but also cannot be greater than 8: 5 ≤ h ≤ 8 is the answer.

 

Hi. Can you explain test 6, section 4, number 14. Thanks.

Sure. The table tells you that the sunflower’s height was 36.36 cm on day 14 and 131.00 cm on day 35. The equation that models that interval well should give you about 36 cm when you substitute 14 for t and give you about 131 cm when you substitue 35 for t. Therefore, you can try each choice using those numbers until you find the one that works.

A) h = 2.1(14) – 15 = 14.4 <– Nope, too small!

B) h = 4.5(14) – 27 = 36 <– Looks good, better check the other value.
h = 4.5(35) – 27 = 130.5 <– Oh yeah, that’s pretty darn close. That’s gotta be our answer.

The other way to go here is to calculate the slope between the two points you know:

\text{slope}=\dfrac{h_2-h_1}{t_2-t_1}=\dfrac{131.00-36.36}{35-14}=4.50666...

The only answer choice with a slope of 4.5 is B, so again, that must be the answer!

Hi Mike,

Can you tell me how to do #27 (Section 4, Practice Test 6). I know how to do it by completing the square but was wondering if there is an easier way.

thanks!
Rachel

If you have a calculator that can graph the given equation without rearranging it (like an Nspire) then the easiest way is just to do that.

       

If you don’t have one of those, though, then completing the square is 100% the way to go.

    \begin{align*}2x^2-6x+2y^2+2y&=45\\\\x^2-3x+y^2+y&=\dfrac{45}{2}\\\\x^2-3x+\dfrac{9}{4}+y^2+y+\dfrac{1}{4}&=\dfrac{45}{2}+\dfrac{9}{4}+\dfrac{1}{4}\\\\\left(x-\dfrac{3}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2&=25\end{align*}

Of course, since the formula for a circle is (x-h)^2+(y-k)^2=r^2, that 25 on the right side of the equation above tells you the radius is 5.

Would you please explain the solution to Problem #31 in the calculator-allowed section of Practice Test #6?

This one is all about translating carefully between words and math. You can write two equations. Let’s use f for the original number of friends. If all f friends go on the trip, the cost per person is $20 less than if f-2 friends go on the trip. Cost per Person with f-2 Friends minus Cost per Person with f Friends is $20.

    \begin{align*}\dfrac{800}{f-2}-\dfrac{800}{f}&=20\end{align}

To solve, multiply by f(f-2):

    \begin{align*}f(f-2)\left(\dfrac{800}{f-2}\right)-f(f-2)\left(\dfrac{800}{f}\right)&=20f(f-2)\\800f-800(f-2)&=20f^2-40f\\800f-800f+1600&=20f^2-40f\\0&=20f^2-40f-1600\\0&=f^2-2f-80\\0&=(f-10)(f+8)\end{align}

That tells you that the number of friends originally in the group is either 10 or –8. Since the answer can’t be negative, it must be 10.

BONUS: The other way to go on this one is the way I actually did it the first time: just dividing 800 by a bunch of numbers in a row until I saw a difference of 20:

800/4 = 200
800/5 = 160
800/6 = 133.333
800/7 = 114.286
800/8 = 100
800/9 = 88.889
800/10 = 80

There you have it: 100 and 80 are 20 apart, and that’s the difference between splitting 800 over 10 people vs. over 8 people. That was fast!

Test 6 calc section number 28

There are two equally valid ways to go here. If you’ve read enough of me, though, you won’t be surprised to learn that I’ll suggest the simpler, less mathy way first.

The question tells you that two points are both 3 units away from –4. You don’t need to make any equations to know what those points are! They’re –1 and –7! So just work through the answer choices until you find one that is true for both x = –1 and x = –7. Start with choice A:

    \begin{align*}|x+4|&=3\\\\|-1+4|&=3\\|3|&=3\\3&=3\\\\|-7+4|&=3\\|-3|&=3\\3&=3\end{align*}

Lookit that: already done!

Algebraically, what’s happening here is that you’re looking to write an equation for “the positive difference between x and –4 is 3.” The way you write “the positive difference” algebraically when you don’t know which of the values you’re subtracting is bigger is you put the subtraction in absolute value brackets. Therefore, “the positive difference between x and –4″ is written |x-(-4)|=|x+4|.

The reason I love the simpler way on this question is that you can just as easily write |-4-x| for the positive difference between x and –4. That’s equivalent because of the absolute value brackets, of course, but if that’s what you start out writing, you might not recognize the correct answer choice right away. If, on the other hand, you just start out by finding the answer choice that works for –7 and –1, you can’t go wrong!

Test 6 calc section number 35

Remember that at the x-intercept, y = 0. Therefore, all you need to do to find the x-intercept is substitute 0 for y and solve!

    \begin{align*}\dfrac{4}{5}x+\dfrac{1}{3}y&=1\\\dfrac{4}{5}x+\dfrac{1}{3}(0)&=1\\\dfrac{4}{5}x&=1\\4x&=5\\x&=\dfrac{5}{4}\end{align*}

Can you do Test 6 #36 from the calculator section

This question rocks!

Since you have data for all the rocks Andrew collected, calculate the average mass of Andrew’s rocks.

\dfrac{2.4+2.5+3.6+3.1+2.5+2.7}{6}=2.8

If the average mass of Maria’s rocks is 0.1 kg greater, then, Maria’s rocks must have an average mass of 2.9 kg.

Now, as in most average questions, you’re going to need to deal with sums. If Maria has 6 rocks, with an average mass of 2.9 kg, then the total mass of her rocks is 6\times 2.9=17.4 kg. Use that to solve for x:

x+3.1+2.7+2.9+3.3+2.8=17.4

x=2.6

Can you do Test 6 section 4 number 37? I always miss this kind of question!

The best way to be sure you’re not missing the mark on a question like this is to actually list the years rather than just trying to capture everything in an equation. So say he starts with x dollars on January 1, 2001. You know his money doubles each year, so you can write the following.

January 1, 2001: x
January 1, 2002: 2x
January 1, 2003: 2(2x) = 4x
January 1, 2004: 2(4x) = 8x
January 1, 2005: 2(8x) = 16x

Of course, you know from the question that on January 1, 2005, Jeremy really had $480 in his account. Easy enough to solve from there:

16x = 480
x = 30

The unnecessarily complicated way to do this question is with an exponent formula. You know his money doubles every year for 4 years, so you can write:

480=x(2^4)

Of course, that just simplifies to the same equation we just solved: 16x = 480.

Test 6 calculator section #34 please!

I’ll give you two good ways to go on this one. First, the algebra. If you know the graph of y=3x^2-14x goes through the point (a,a), then you can substitute a in for x and y and solve!

    \begin{align*}a&=3a^2-14a\\0&=3a^2-15a\\0&=a^2-5a\\0&=a(a-5)\end{align*}

From that, you know a must equal 5, since the question distinguishes between (0,0) and (a,a).

The other way to go is to use your graphing calculator—just find the intersections between y=3x^2-14x and y=x.

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