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2 years ago

5 seniors, 2 juniors, 6 sophomores, 5 fresh men's. What is the ratio of sophomores to all bus riders?

Mathematics

2 answers:

tatiyna2 years ago

3 0

I think the answer is 6:18

photoshop1234 [79]2 years ago

3 0

18:6 sophomores in the bus

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Cound someone please help me on this, I'm stuck.

Katyanochek1 [597]

Negative exponents work like this:

$a^{-b}=\dfrac{1}{a^b}$

So, in order to evaluate a negative exponent, you simply have to invert the base, and then raise to the positive equivalent of the exponent.

As an example, here are the first three exercises:

$8^{-3}=\dfrac{1}{8^3}=\dfrac{1}{512}$

$(-4)^{-5}=\dfrac{1}{(-4)^5}=-\dfrac{1}{1024}$

$2k^{-4}=\dfrac{2}{k^4}$

You can work out the rest applying this logic.

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3 years ago

Travis bought 8 new books for $7.99 each. How much did he spend on books?

Otrada [13]

Answer:

$63.92

Step-by-step explanation:

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2 years ago

Bruce's hourly wage increased from $15.50 to $18.60.

Serga [27]

Answer:

$1.20 Increase

Step-by-step explanation:

18.60/15.5

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2 years ago

Calculus piecewise function.

Kipish [7]

Part A

The notation $\lim_{x \to 2^{+}}f(x)$ means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.

So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc

We don't actually arrive at x = 2 itself. We simply move closer and closer.

Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2

Therefore f(x) = (x/2) + 1

Plug in x = 2 to find that...

f(x) = (x/2) + 1

f(2) = (2/2) + 1

f(2) = 2

This shows $\lim_{x \to 2^{+}}f(x) = 2$

Then for the left hand limit $\lim_{x \to 2^{-}}f(x)$ , we'll involve x < 2 and we go for the first piece. So,

f(x) = 3-x

f(2) = 3-2

f(2) = 1

Therefore, $\lim_{x \to 2^{-}}f(x) = 1$

===============================================================

Part B

Because $\lim_{x \to 2^{+}}f(x) \ne \lim_{x \to 2^{-}}f(x)$ this means that the limit $\lim_{x \to 2}f(x)$ does not exist.

If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.

===============================================================

Part C

You'll follow the same type of steps shown in part A.

However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.

So you'd only involve the second piece f(x) = (x/2) + 1

You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.

===============================================================

Part D

As mentioned earlier, since $\lim_{x \to 4^{+}}f(x) = \lim_{x \to 4^{-}}f(x) = 3$ , this means the limit $\lim_{x \to 4}f(x)$ does exist and it's equal to 3.