Answer:
<h2>(g-f)(10) = - 71</h2>
Step-by-step explanation:
f(x) = x² - 1
g(x) = 2x + 8
To find (g-f)(10) first find ( g - f)(x)
To find ( g - f)(x) subtract f(x) from g(x)
That's
( g - f)(x) = 2x + 8 - ( x² - 1)
Remove the bracket
( g - f)(x) = 2x + 8 - x² + 1
Simplify
( g - f)(x) = - x² + 2x + 9
To find (g-f)(10) substitute the value in the bracket that's 10 into ( g - f)(x)
That is
(g-f)(10) = -(10)² + 2(10) + 9
= - 100 + 20 + 9
= - 100 + 29
= - 71
Hope this helps you
Percent Change = New Value − Old Value|Old Value| × 100%
Example: There were 200 customers yesterday, and 240 today:
240 − 200|200|× 100% = 40200 × 100% = 20%
A 20% increase.
Percent Error = |Approximate Value − Exact Value||Exact Value| × 100%
Example: I thought 70 people would turn up to the concert, but in fact 80 did!
|70 − 80||80| × 100% = 1080 × 100% = 12.5%
I was in error by 12.5%
(Without using the absolute value, the error is −12.5%, meaning I under-estimated the value)
The difference between the two is that one states factual calculations and the other is a theoretical guess
Answer:
Step-by-step explanation:
16 x 28% = 4.48
16 x 72% = 11.52
16 x 172% = 27.52
16 x 128% = 20.48
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I dont know for the second one, sorry-
<h3>
Answer: 375</h3>
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Work Shown:
a = 300 = first term
r = 60/300 = 0.2 = common ratio
We multiply each term by 0.2, aka 1/5, to get the next term.
Since -1 < r < 1 is true, we can use the infinite geometric sum formula below
S = a/(1-r)
S = 300/(1-0.2)
S = 300/0.8
S = 375
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As a sort of "check", we can add up partial sums like so
- 300+60 = 360
- 300+60+12 = 360+12 = 372
- 300+60+12+2.4 = 372+2.4 = 374.4
- 300+60+12+2.4+0.48 = 374.4+0.48 = 374.88
and so on. The idea is that each time we add on a new term, we should be getting closer and closer to 375. I put "check" in quotation marks because it's probably not the rigorous of checks possible. But it may give a good idea of what's going on.
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Side note: If the common ratio r was either r < -1 or r > 1, then the terms we add on would get larger and larger. This would mean we don't approach a single finite value with the infinite sum.
