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2 years ago

If the temperature started at 8 degrees celcius at Noon and it dropped 10 degrees over the next 5 hours

Mathematics

1 answer:

leonid [27]2 years ago

8 0

Answer:

The answer is " $\bold{2^{\circ}\ C}$ "

Step-by-step explanation:

Given:

noon temperature $= 8^{\circ} C$

Total change in next 5 hours $= 10^{\circ} C$

Calculating the final temperature:

$\text{final temperature = noon temperature - changes in 5 hours}$

$= 8^{\circ} -10^{\circ} \\\\=- 2^{\circ} \ C$

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Students took a typing test. The number of words per minute students could type are listed below. 40, 47, 25, 32, 48, 55, 33, 56

const2013 [10]

Answer:

Step-by-step explanation:

The median number is known as the middles number

The first step is to rearrange the set of numbers starting from the lowest

25, 32, 32, 32, 33, 40, 47, 47, 48, 55, 56

Hence the middle number is 50

8 0

2 years ago

The eleventh term from the end of the AP 14, 19, 24, …………., 264 is

vazorg [7]

Answer:

<h2>The eleventh term of the sequence is 64</h2>

Step-by-step explanation:

The sequence given is an arithmetic sequence

14, 19, 24, …………., 264

The nth term of an arithmetic sequence is given as;

Tn = a+(n-1)d where;

a is the first term = 14

d is the common difference = 19-14=24-19 = 5

n is the number of terms = 11(since we are to look for the eleventh term of the sequence)

substituting the given values in the formula given;

T11 = 14+(11-1)*5

T11 = 14+10(5)

T11 = 14+50

T11 = 64

The eleventh term of the sequence is 64

6 0

3 years ago

The table shows locations on a map.

Nina [5.8K]

Answer:

The table shows locations on a map.

Place

park
farmers market
post office
bus stop

Location

(-2,2)
(-2,3)
(2, 2)
(6,-3)

Which places on the map are 4 units apart?

O the park and the farmer's market

O the farmers market and the bus stop

✓ the post office and the park

O the post office and the bus stop

Step-by-step explanation:

Distance :---

$→ \sqrt{ {(2 - ( -2 ))}^{2} + {(2 - 2)}^{2} } \\ = \sqrt{ {4}^{2} + {0}^{2} } = \sqrt{ {4}^{2} } = \boxed{4 \: units}$

5 0

3 years ago

WHICH IS NOT A LINEAR FUNCTION

Tom [10]

Answer:

OPTION C

Step-by-step explanation:

IN OPT. A- 2 IS MULTIPLED IN EACH NO.

IN OPT.B-2 IS MULTIPLED

IN OPT.D-1 IS ADDED

IN OPT.E-2 IS ADDED

4 0

2 years ago

A 100 gallon tank initially contains 100 gallons of sugar water at a concentration of 0.25 pounds of sugar per gallon suppose th

Vsevolod [243]

At the start, the tank contains

(0.25 lb/gal) * (100 gal) = 25 lb

of sugar. Let $S(t)$ be the amount of sugar in the tank at time $t$ . Then $S(0)=25$ .

Sugar is added to the tank at a rate of P lb/min, and removed at a rate of

$\left(1\frac{\rm gal}{\rm min}\right)\left(\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm gal}\right)=\dfrac{S(t)}{100}\dfrac{\rm lb}{\rm min}$

and so the amount of sugar in the tank changes at a net rate according to the separable differential equation,

$\dfrac{\mathrm dS}{\mathrm dt}=P-\dfrac S{100}$

Separate variables, integrate, and solve for S.

$\dfrac{\mathrm dS}{P-\frac S{100}}=\mathrm dt$

$\displaystyle\int\dfrac{\mathrm dS}{P-\frac S{100}}=\int\mathrm dt$

$-100\ln\left|P-\dfrac S{100}\right|=t+C$

$\ln\left|P-\dfrac S{100}\right|=-100t-100C=C-100t$

$P-\dfrac S{100}=e^{C-100t}=e^Ce^{-100t}=Ce^{-100t}$

$\dfrac S{100}=P-Ce^{-100t}$

$S(t)=100P-100Ce^{-100t}=100P-Ce^{-100t}$

Use the initial value to solve for C :

$S(0)=25\implies 25=100P-C\implies C=100P-25$

$\implies S(t)=100P-(100P-25)e^{-100t}$

The solution is being drained at a constant rate of 1 gal/min; there will be 5 gal of solution remaining after time

$1000\,\mathrm{gal}+\left(-1\dfrac{\rm gal}{\rm min}\right)t=5\,\mathrm{gal}\implies t=995\,\mathrm{min}$

has passed. At this time, we want the tank to contain

(0.5 lb/gal) * (5 gal) = 2.5 lb

of sugar, so we pick P such that

$S(995)=100P-(100P-25)e^{-99,500}=2.5\implies\boxed{P\approx0.025}$

5 0

3 years ago