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3 years ago

The sum of two numbers is 61 . The larger number is 5 more than the smaller number. What are the numbers?

Mathematics

1 answer:

Elanso [62]3 years ago

6 0

Answer:

35.5 and 25.5

Step-by-step explanation:

all you have to do is take 61÷2=30.5 and you add five to one of the halves, and subtract five form the other.

61÷2= 30.5

30.5+5= 35.5

30.5-5= 25.5

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3 years ago

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1) A cyclist travels a distance of 90 miles in 5 hours. What

Nostrana [21]

Answer:

Step-by-step explanation: Speed is distance an object travelled per unit time.

It is represented by letter v and can assume various units.

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v = 18 miles per hour

Her average speed is 18 mph.

2) Rearraging formula for distance:

d = v.t

d = 70 * 4

d = 280 miles

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$v=\frac{315}{6}$

v = 52.5 mph

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$t=\frac{d}{v}$

$t=\frac{210}{90}$

t = 2.34 hours

It will take 2 hours and half to arrive in Leeds. So, we will be expected to arrive at 12:04.

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$t=\frac{2000}{4}$

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or t = 8.34 minutes

It will take, for an athlete to run 2000m, 8.34 minutes.

6) a) All units must match, so for a speed in mph:

45 minutes is 0.75 hours. Total time is 8.75 hours.

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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,

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Separate variables, integrate, and solve for S.

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

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

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

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Use the initial value to solve for C :

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

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

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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

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

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2 years ago

I need help on this math problem.

OLEGan [10]

B: (0, a)

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2 years ago

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Answer:

Step-by-step explanation:

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