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nignag [31]
3 years ago
5

Ashley wants to join a cooking class. She is trying to decide which membership will be the best buy after 10 lessons. Which clas

s is the better price for 10 lessons?
Class 1 - $20 sign up fee, $20 per lesson

Class 2 - $40 sign up fee, $12 per lesson

Answer 1) Class 1
Answer 2) Class 2
Mathematics
2 answers:
atroni [7]3 years ago
8 0
Hey! Let's help you with that.

When we are looking at Class 1. There is a $20 sign up fee which we will have to add to our total. If it is $20 per lesson and we are finding the better price for 10 lessons, we can take the $20 and multiply it by the 10 lessons she wants to take. Which would give us $200. Add on the $20 signup fee and the total becomes $220 for Class 1.

When it comes to Class 2, we have a steeper sign up fee of $40, however it is $12 per lesson. We simply do the exact same thing as we did in Class 1, where we take $12 and multiply it by the 10 lessons to get $120 and then add on the signup fee which would total up to $160.

So in the end, Class 2 would be the better price for 10 lessons as the total cost of it would only be $160, compared to the Class 1 total price which is $220.
Aleksandr-060686 [28]3 years ago
5 0
Class one will cost $220 while Class two will cost $160.
Therefor the answer is answer 2
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3 years ago
Able, Ben and Cal each played a game able’s score was six times ben’s score, Cale score was a hird time of Able’s score write do
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Answer:

6:1:2

Step-by-step explanation:

Let a = Able's score, b = Ben's score, and c = Cal's score.

Since

Able's score was 6 times Ben's score, that means a = 6b.

Cal's score was a third of Able's score, so that means c = a/3. And since a = 6b, that means c = 6b / 3 = 2b.

Thus, the ratio of Able's score to Ben's score to Cal's score, a:b:c, is 6:1:2, because c is twice as much as b and a is 6 times as much as b.

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3 years ago
in Oklahoma the states sales tax is 7%.if James purchases a lawn mower then how much will James pay in states sales tax​
Nimfa-mama [501]

Answer:

Step-by-step explanation:

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7 0
3 years ago
Can someone help me with this? I need to find the points of discontinuity/limits for each of these. I think one point is 4, but
Debora [2.8K]
The answers are shown in the attached image

-------------------------------------------------------------------------

Explanation:

Set the denominator x^4-8x^3+16x^2 equal to zero and solve for x

x^4-8x^3+16x^2 = 0
x^2(x^2-8x+16) = 0
x^2(x-4)^2 = 0
x^2 = 0 or (x-4)^2 = 0
x = 0 or x-4 = 0
x = 0 or x = 4

The x values 0 and 4 make the denominator zero

These x values lead to asymptote discontinuities because the numerator 8x-24 = 8(x-3) has no common factors which cancel with the denominator factors.

There are two vertical asymptotes

Let's see what happens when we plug in a value to the left of x = 0, say x = -1, we'd get
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(-1) = (8(-1)-24)/((-1)^4-8(-1)^3+16(-1)^2)
f(-1) = -1.28
So as x gets closer and closer to x = 0 from the left side, the f(x) is heading to negative infinity

Now plug in some value to the right of x = 0. I'm going to pick x = 1
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(1) = (8(1)-24)/((1)^4-8(1)^3+16(1)^2)
f(1) = -1.78 (approximate)
So as x gets closer and closer to x = 0 from the right side, the f(x) is heading to negative infinity

Overall, as x approaches 0 from either the left or right side of x = 0, the y value is heading off to negative infinity

---------------------

Repeat for values to the left and right of x = 4
We can't use x = 1 as it turns out that x = 3 is a root
But we can use something like x = 3.5 to find that...
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(3.5) = (8(3.5)-24)/((3.5)^4-8(3.5)^3+16(3.5)^2)
f(3.5) = 1.31 approx
So as x gets closer to x = 4 from the left, y is getting closer to positive infinity

Plug in x = 5 to find that
f(x) = (8x-24)/(x^4-8x^3+16x^2)
f(5) = (8(5)-24)/((5)^4-8(5)^3+16(5)^2)
f(5) = 0.64
which has the same behavior as the left side

So overall, as we approach x = 4, the y value is heading off to positive infinity

Again everything is summarized in the image attachment

Note: you could make a table of more values but they would effectively say what has already been said. It would be redundant busy work. However, its always good practice for function evaluation. 

6 0
3 years ago
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