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jasenka [17]
3 years ago
10

The temperature of a pot of water was 180.3°F and it cooled at a rate of -2.5°F for a minute after 20 minutes what was the tempe

rature
Mathematics
1 answer:
kykrilka [37]3 years ago
7 0
Temperature after 20 minutes is found by multiplying-2.5 by 20 since it goes down by 2.5 degrees for one minute. -2.5 * 20 = -50. Adding this to 180.3 will give the temperature 130.3 degrees.
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Function f(x) is positive, increasing and concave up on the closed interval [a, b]. The interval [a, b] is partitioned into 4 eq
Umnica [9.8K]

The left sum would be f0+f1+f2+f3

The right sum would be f1+f2+f3+f4

The trapezoidal rule value is:

(f0+f1)/2 + (f1+f2)/2+(f2+f3)/2 +(f3+f4)/2

This would put the trapezoidal rule in the middle , which makes the answer:

Lower sum < Trapezoidal rule Value < Upper sum

5 0
3 years ago
Read 2 more answers
Can someone please help me .?
MariettaO [177]

Answer:

5

Step-by-step explanation:

5+7

4 0
3 years ago
Ann car can go 117 miles on 3 gallons of gas. During a drive last weekend, Ann used 8 gallons of gas. How far did she drive?
Nitella [24]

Answer:

Ann drove 312 miles in 8 gallons.

Step-by-step explanation:

Distance covered by Ann's car = 117 miles

Gallons of gas spent by Ann's car to cover 117 miles = 3 gallons

Unit rate = 117 miles / 3 gallons

               = 39 miles per gallon

so Ann's car can go 39 miles in one gallon of gas.

  • Given that Ann used 8 gallons of gas.

Therefore,

Distance covered by Ann's car in 8 gallons = 39 × 8

                                                                         = 312 miles

Therefore, Ann drove 312 miles in 8 gallons.

6 0
3 years ago
B. Determine the values of a and b so that f is continuous.<br><br> Please help!
wlad13 [49]

Answer:

following are the solution to this question:

Step-by-step explanation:

Please find the complete question in the attached file.

\lim_{x\to 2}+f(x) = \lim_{x\to 2}+ (ax^2-bx+3) = 4a-2b+3\\\\ \to  4a-2b+3 = 4\\\\  \therefore \ \ 4a-2b = 1\\\\

The one-sided limits of F(x) at x = 3 must be equivalent for f(x) to be continuous at x = 3.

\to \lim_{x\to 3}- f(x) = \lim_{x\to 3}- (ax^2-bx+3) = 9a-3b+3\\\\ \to \lim_{x\to 3}+ f(x) = \lim_{x\to 3}+ (2x-a+b) = 6-a+b\\\\  

So,

\to 9a-3b+3 = 6-a+b\\\\\therefore\\ \to 10a -4b = 3

\to 4a - 2b = 1......(a)\\\to 10a - 4b = 3.......(b)\\

In equation a multiply the by -2 and then add in the equation b:

\to -8a + 4b = -2\\\to 10a - 4b = 3\\ \to 2a = 1\\\\ \to   a = \frac{1}{2}\\\\ \to \ 4(\frac{1}{2}) - 2b = 1\\\\ \to 2 - 2b = 1\\\\ \to   -2b = -1\\\\ \to b = \frac{1}{2}  

So, the value of a \ and \ b= \frac{1}{2}

4 0
3 years ago
9 students volunteer for a committee. How many different 2-person committees can be chosen?
asambeis [7]
It is definitely going to be 72

7 0
3 years ago
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