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KIM [24]
3 years ago
9

C/-3+10=9 that is what I need help on

Mathematics
2 answers:
Digiron [165]3 years ago
7 0
I think the answer is C=63
Elden [556K]3 years ago
3 0

Answer:

i think it is -3

Step-by-step explanation:

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The number of gallons of water in a large tank at time (t) minutes is given by the function W(t)=160,000-8,000t+t². Find the ave
sergij07 [2.7K]

The number of gallons of water in the tank at t=10 is

... W(10) = 160,000 -10(8000 -10) = 80100

The number of gallons of water in the tank at t=10.5 is

... W(10.5) = 160,000 -10.5(8000 -10.5) = 76110.25

The rate of change over the interval is

... (W(10.5) - W(10))/(10.5 - 10) = (76110.25 - 80100)/(0.5) = -7979.5

The average rate of change in the number of gallons of water in the tank over the interval is -7979.5 gal/min.

The sign is negative, so the amount of water is decreasing.

3 0
3 years ago
What value of a makes the statement below true?
mr Goodwill [35]

Answer:

The answer would be D, 75.

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
Need help question 11
Helga [31]
I am sorry i dont understand

5 0
3 years ago
Jimmy is selling bracelets for $1.50. So far she has earned $120. She needs to earn more than $135 in order to meet her goal. Ho
Tema [17]
Selling 90 would reach her goal of $135 so she would need to sell 91 to exceed her goal 

135/1.5=90 so 91*1.5=136.5
4 0
3 years ago
Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

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

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
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