Posts tagged with: Tumblr

The midpoint formula tells you that the a segment with endpoints (a, b) and (c, d) will have a midpoint at ((c)/2,(d)/2)

So we know that (+ 9)/2 = 6 and (5 + y)/2 = 3. We can solve those!

(+ 9)/2 = 6
+ 9 = 12
x = 3

(5 + y)/2 = 3
5 + y = 6
y = 1

Therefore, x + y = 4.

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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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Just do those things and simplify:

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Looks like the volume of the second cylinder is half the volume of the first! 🙂

The shortcut here is to recognize that the r value gets squared but the h value doesn’t, so you can say the volume is divided by 4 then multiplied by 2.

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60° is 1/6 of the circle (which has 360° of arc in total), so the length of the minor arc will be 1/6 of the circumference. 1/6 of 12π is 2π.

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Huge shortcut here if you just know that for a parabola in standard ax^2 + bx + c form, the x-coordinate of the vertex will be at –b/(2a). In this case, that means it’s at –3/(2(–6)) = 3/12 = ¼.

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The way to think about this (for me, anyway) begins with understanding that 2[something] + 3 = 8x – 1, and our job is to figure out what that something is.

Since this question gives us answer choices, all we really need to do is try each one as the something to see what works. Since they’re not in numerical order, start with A.

2(4x – 2) + 3
8x – 4 + 3
8x – 1

Oooh, look at that! we’re already done! 🙂

(Also, for those worrying along at home, note that 5 answer choices means this isn’t an SAT question. Could be SAT Subject Test, though.)

If you didn’t have answer choices, you’d still start in the same place, but then think through logically. Thought process:

2[something] + 3 = 8x – 1

Hmm..there’s gotta be a 4x in the something otherwise i’ll never get 8x. But if it’s just 4x in there then I’ll end up with 8x + 3, not 8x – 1. How do i take 4 away? By adding a –2 inside the something!

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When you have exponents with the same base and you divide them, you subtract the exponents.

So from what we’re given, we know that 2x – (xy) = 1. We can solve that for y.

2x – (xy) = 1
2x – x + y = 1
x + y = 1
y = 1 – x

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Start by drawing it!

Note that OC = 5 and OD = 5 because both of those are also radii. Note also that because chord CD is perpendicular to OB, it’s bisected by OB. In other words, it’s split into 2 segments each measuring 4.

 

Things are really coming together! Because we know our Pythagorean triples when we see them, we know that we’ve got 3-4-5 triangles here. Therefore, the sine of angle OCD is 3/5.

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One of the two things being multiplied must be zero, so either x – 3 = 0 or x + 7 = 0.

If x + 7 = 0, then x = –7. In that case, we can substitute into x – 3:

x – 3
= –7 – 3
= –10

So x – 3 can equal 0 or –10.

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Draw!

You know point C must be on the x-axis, and that AB = BC. Since B is on the y-axis, it turns out all we need to do here is reflect point A across the y-axis. Point C must be at (8, 0), so x = 8.

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The correct answer is A. Subtract c from both sides of the inequality and you’re left with 8 > 4. That’s always true!

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In 200 ml of 30% saline, there’s (0.3)(200) = 60 ml of saline.

60 ml is 40% of what?

60 = (0.4)x
150 = x

So the saline will be 40% when enough water evaporates that there’s 150 ml of solution. Therefore, 50 ml of the original 200 ml must evaporate.

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Let’s work backwards, and use hours instead of distance for ease (assume the plane travels at a constant speed for our purposes).

Let’s say when she woke up, she had 1 hour left in her flight. That’s half of the time she was sleeping so she must’ve slept for 2 hours.

She first fell asleep halfway through the flight, so she was awake for the first 3 hours, then slept for 2, then was awake for the last 1.

She was asleep for 2/6 = 1/3 of the trip.

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Rearrange that equation:

x + 2y = y
x = –y
x/y = –1

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Note that the SAT doesn’t test properties of even and odd like this, although the old SAT (pre-2016) used to.

The product of two numbers will be even if and only if one or both of the numbers being multiplied is even. Therefore, you know any card showing an odd number must have an even number on the other side.

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