Posts tagged with: lines

When a system of linear equations has no solution, that means you have parallel lines, which means the lines have the same slope. So put both equations into slope-intercept form (y = mx + b) first:

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In order for those lines to be parallel, their slopes must be equal, which means 2/5 = -4/k. That means k must be equal to –10.

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Trigonometry does the trick here. Below is that line making a 42° angle with the positive x-axis. I’ve also drawn a dotted segment to make myself a neat little right triangle.

Remember that slope is rise over run—how high the line climbs divided by how far it travels right. In this case, the dotted segment labeled a is the rise and the bottom of the triangle labeled b is the run. And luckily for us, the tangent function calculates that a/b ratio! Remember your SOH-CAH-TOA. Tangent = Opposite/Adjacent.

Just use your calculator to evaluate tan 42°. You’ll get 0.90.

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Hi Mike,

Here is a really confusing question from Applerouth’s SAT text:

a = 1.5 x + 1.50
b = 1.25x + 4.50

In the system of equations above, a and b represent the cost, in dollars, of buying x buffalo wings at two different restaurants. What amount of money will get you the same number of buffalo wings at both restaurants?

A) 12
B) 19.5
C) 20
D) 29.5

The answer is A. No idea how to do this.

You have to find this by looking for the number of wings that costs the same at both stores, so set a and b (the costs at each store) equal to each other and then solve for (the number of wings that will make each store’s cost the same).

1.5x + 1.50 = 1.25x + 4.50
0.25x = 3
x = 12

Therefore, buying 12 wings at each store costs the same amount of money. The question appears to ask HOW MUCH money, so to finish the problem you need to plug 12 back in for x in either of the equations. I’ll do the first one:

a = 1.5(12) + 1.50
a = 19.50

So the answer really should be B. The answer would be A if the question asked for the number of wings that cost the same at both stores, but that’s not what the question asks.

One other note: it might make it more clear what’s going on to graph each line. What the graph below shows is that the price (on the y-axis) is cheaper at store a for up to 12 wings, but store b becomes a better deal for 13 or more wings. At the intersection point—12 wings, $19.50—the same number of wings costs the same amount at both stores.

Can you work out problem #11 of section four in Official SAT #5?

Sure. The math concept you need to know here is that perpendicular lines have negative reciprocal slopes. For example, if one line has a slope of \frac{7}{2}, then a line perpendicular to it must have a slope of -\frac{2}{7}.

In this case, none of the lines are given to you in slope-intercept form, so your first step should be to put at least the first line you’re given into that form.

    \begin{align*}-2x+3y&=6\\3y&=2x+6\\y&=\frac{2}{3}x+2\end{align*}

That tells you that the slope of the given line is \frac{2}{3}, so the slope you need to find in the answer choices is -\frac{3}{2}.

They do you a real favor in the answer choices by only using 3 and 2 as coefficients together in one choice. So just from the get-go you should be leaning choice A. Let’s just put it in slope-intercept form to be sure.

    \begin{align*}3x+2y&=6\\2y&=-3x+6\\y&=-\frac{3}{2}x+3\end{align*}

Yep, that’s what we wanted. We’re all done!

Note that f(x) is a line in slope-intercept form, where a is the slope and b is the intercept.

Once you recognize that, you just need to know that, notationally, the –2 in front of f(x) means you multiply the whole thing by –2. So:

image

That’s still a line, but now the slope is –2a.

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Note that f(x) is a line in slope-intercept form, where a is the slope and b is the intercept.

Once you recognize that, you just need to know that, notationally, the –2 in front of f(x) means you multiply the whole thing by –2. So:

image

That’s still a line, but now the slope is –2a.

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Hi. Would it be possible for you to explain #13 on SAT practice test 5 on calculator inactive (the tea bag problem)? I think the main reason I found it confusing was the wording, and the SAT explanation for the answer was also a little bit wordy.
Thanks

I’ll try to be as un-wordy as possible! 🙂

I think plugging in numbers helps here. Say the restaurant wants to make 20 cups of tea one night, so n=20. The equation t=n+2 tells us that they’d use t=22 tea bags to make 20 cups of tea.

Now, what if they wanted to make one more cup of tea (in our example, 21 cups instead of 20 cups)? Well, the equation tells us they’d use 23 tea bags to do that:

    \begin{align*}t&=n+2\\t&=21+2\\t&=23\end{align*}

So they needed 22 bags to make 20 cups, and 23 bags to make 21 cups. That’s one extra bag for one extra cup.

The other way to think about this is that they’re just giving you a linear equation and asking for the rate of change (i.e., the slope). Think of the equation t=n+2 as y=x+2. That’s a line with a slope of 1 (and a y-intercept of 2, but we don’t need that for this question). A slope of 1 means for every change in x, you get an equal change in y. In other words, for every change in cups of tea, you get an equal change in tea bags.

Do either of those explanations help?

Test 8 Section 3 #13

You’re translating from words into math here. Key words to note: “at a constant rate” means you’re dealing with something linear, that has a slope. And “decreased at a constant rate” means you’re looking for a negative slope.

Because the function is a function of t, years are your x-variable and and millions of barrels is your y-variable. Calculate the slope from the numbers the question provides: 4 million barrels in 2000 and 1.9 million barrels in 2013.

    \begin{align*}\text{Slope}&=\dfrac{4-1.9}{2000-2013}\\&=\dfrac{2.1}{-13}\end{align*}

Note that while no answer choice has that slope in that exact form, choice C is equivalent. \dfrac{2.1}{-13}=-\dfrac{21}{130}.

Because all the answer choices have the same y-intercept of 4, you need not think much about it, but it does make sense given the question. There were 4 million barrels in 2000 and t represents years after 2000, so t = 0 when the y-variable is 4.

The graph of a linear function f has a positive slope with intercepts (a,0) and (0,b), where a and b are non-zero integers. Which of the following statements about a and b could be true?

A) a + b = 0

B) a – 2b = 0

C) a = b

D) 0 <a < b

(I only know that Choice C is out because that would be true only if the slope=1 and the line passed through the origin, but since a and b are non-zero integers, there can be no point (0,0), so that one answer choice is out. )

It helps to do a bit of drawing here. Lines with positive slopes (that don’t pass through the origin) will always have either a positive y-intercept and a negative x-intercept, or a negative y-intercept and a positive x-intercept.

Therefore, you can eliminate choices B, C ,and D because in those answers, nonzero integers a and b must be the same sign. (Choice D goes further than B and C, requiring a and b to both be positive.) Choice A works because if a and b are nonzero, then they must have opposite signs for ab = 0.

Test 5 Section 4 Number 23

All we really need to do to nail this question is match a table to a linear function. Unlike some similar questions, the table isn’t labeled with variables and the values aren’t sorted in ascending or descending order, but those are just details! We don’t need to get too creative here; let’s just backsolve through the choices and see which one works best. Remember that r represents the monthly rental price (the rightmost column) and p represents the purchase price in thousands (the middle column).

Let’s drop the first row of values (Clearwater Lane) into choice A:

    \begin{align*}r(p)=2.5p-870\\950=2.5(128)-870\\950=-550\end{align*}

NOPE! That didn’t work. We can eliminate A. Let’s try B:

    \begin{align*}r(p)=5p+165\\950=5(128)+165\\950=805\end{align*}

That’s not quite as ridiculous, but still isn’t close enough. Let’s try C:

    \begin{align*}r(p)=6.5p+440\\950=6.5(128)+440\\950=1275\end{align*}

No way, that’s not even close. Eliminate C. Try D:

    \begin{align*}r(p)=7.5p-10\\950=7.5(128)-10\\950=950\end{align*}

Huh. That’s pretty good! To make sure D is the answer, let’s try a couple other rows of data: Driftwood Drive and Edgemont Street.

    \begin{align*}r(p)=7.5p-10\\1310=7.5(176)-10\\1310=1310\end{align*}

    \begin{align*}r(p)=7.5p-10\\515=7.5(70)-10\\515=515\end{align*}

Both are perfect! That’s enough to satisfy me. D is the answer.

Test 3 Section 4 #26

Fun fact and major shortcut: when a line passes through the origin, the coordinates of each point are in the same ratio. For example, if you knew a line passed through (3, 6) and (30, 60), you’d know that line passed through the origin. That’s because when a line passes through the origin, its equation will simply be ymx + 0, or simply ymx. Put another way, y will be directly proportional to x.

We can use that fact to solve quickly:

    \begin{align*}\frac{2}{k}&=\frac{k}{32}\\\\k^2&=64\\k&=\pm 8\end{align*}

Of course, because there are no negative answer choices, we choose C for 8.

The other way to go here would be to backsolve by just trying to draw the points. You can cross off A right away because any line that goes through (2, 0) and (0, 32) can’t also go through the origin! Even a rough drawing should convince you that the line containing(2, 16) and (16, 32) doesn’t go through the origin. Things get a little trickier with (2, 4) and (4, 32) because it might look CLOSE, but hopefully the right answer looks even closer.

Test 1 Calculator OK #16

The graph here is definitely a linear relationship between h and C, so we’re going to want to figure out the slope and the C-intercept. Then we’ll use the standard y=mx+b slope-intercept formula, only for this problem it’ll be C=mh+b.

Slope is rise over run, or the change in the vertical variable divided by the change in the horizontal variable. Let’s pick two points from the line and plug them into the slope formula. How about (1, 8) and (3, 14)?

    \begin{align*}\text{slope}=\dfrac{C_2-C_1}{h_2-h_1}=\dfrac{14-8}{3-1}=3\end{align*}

Once we have the slope, we’re actually done because whoever wrote this question wasn’t very tricky with the answer choices. The C-intercept of this graph is obviously not zero, and only choice C has the slope we just calculated plus a non-zero C-intercept. Sure enough, a quick glance at the graph confirms that the C-intercept is 5, meaning the equation should be C=3h+5.