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2 years ago

Work out the value of each expression, when r= 2.5 and t = 6.

Mathematics

2 answers:

Usimov [2.4K]2 years ago

8 0

Answer:

a) 4(t - r) = 4(6 - 2.5) = 4 x 3.5 = 14

b) 4t - r = (4 x 6) - 2.5 = 24 - 2.5 = 21.5

c) 2(3t - 10) = 2(3 x 6 - 10) = 2(18 - 10) = 2 x 8 = 16

d) (t - 2)² = (6 - 2)² = 4² = 4 x 4 = 16

e) cannot decipher the equation

f) 6r + t = 6 x 2.5 + 6 = 15 + 6 = 21

bonufazy [111]2 years ago

8 0

Answer:

the circumference of wheel is 880cm calculate the diameter

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3 years ago

Read 2 more answers

The temperature, H, in °F, of a cup of coffee t hours after it is set out to cool is given by the following equation. H = 70 + 1

Alex73 [517]

Answer:

The temperature a t = 0 is 190 °F

The temperature a t = 1 is 100 °F

The temperature a t = 2 is 77.5 °F

It takes 1.5 hours to take the coffee to cool down to 85°F

It takes 2.293 hours to take the coffee to cool down to 75°F

Step-by-step explanation:

We know that the temperature in °F, of a cup of coffee t hours after it is set out to cool is given by the following equation:

$H(t)=70+120(\frac{1}{4})^t$

a) To find the temperature a t = 0 you need to replace the time in the equation:

$H(0)=70+120(\frac{1}{4})^0\\H(0)=70+120\cdot 1\\H(0) = 70+120\\H(0)=190 \:\°F$

b) To find the temperature after 1 hour you need to:

$H(1)=70+120(\frac{1}{4})^1\\H(1)=70+120(\frac{1}{4})\\H(1) = 70+30\\H(1)=100 \:\°F$

c) To find the temperature after 2 hours you need to:

$H(2)=70+120(\frac{1}{4})^2\\H(2)=70+120(\frac{1}{16})\\H(2) = 70+\frac{15}{2} \\H(2)=77.5 \:\°F$

d) To find the time to take the coffee to cool down $85 \:\°F$ , you need to:

$85 = 70+120(\frac{1}{4})^t\\70+120\left(\frac{1}{4}\right)^t=85\\70+120\left(\frac{1}{4}\right)^t-70=85-70\\120\left(\frac{1}{4}\right)^t=15\\\frac{120\left(\frac{1}{4}\right)^t}{120}=\frac{15}{120}\\\left(\frac{1}{4}\right)^t=\frac{1}{8}$

$\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(\left(\frac{1}{4}\right)^t\right)=\ln \left(\frac{1}{8}\right)$

$\mathrm{Apply\:log\:rule}=\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$t\ln \left(\frac{1}{4}\right)=\ln \left(\frac{1}{8}\right)$

$t=\frac{\ln \left(\frac{1}{8}\right)}{\ln \left(\frac{1}{4}\right)}\\t=\frac{3}{2} = 1.5 \:hours$

e) To find the time to take the coffee to cool down $75 \:\°F$ , you need to:

$75=70+120\left(\frac{1}{4}\right)^t\\70+120\left(\frac{1}{4}\right)^t=75\\70+120\left(\frac{1}{4}\right)^t-70=75-70\\120\left(\frac{1}{4}\right)^t=5\\\left(\frac{1}{4}\right)^t=\frac{1}{24}$

$\ln \left(\left(\frac{1}{4}\right)^t\right)=\ln \left(\frac{1}{24}\right)\\t\ln \left(\frac{1}{4}\right)=\ln \left(\frac{1}{24}\right)\\t=\frac{\ln \left(24\right)}{2\ln \left(2\right)} \approx = 2.293 \:hours$

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jekas [21]

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