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3 years ago

Write the fact family for these factors: 8, 64

Mathematics

1 answer:

natka813 [3]3 years ago

3 0

Answer:

8*64=512

64*8=512

512/64=8

64/512=8

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60$. These are my equations :
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Then...
600 * 0.1 = 60 ; 600 - 60 = 540
Which then became....
540 - 480 = 60

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During a sale, cameras were discounted 30% from their original price of $295.99. To the nearest cent, how much was the camera du

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What is the solution to the system of equations below?<br> Y=-1/3x+6 and x=-6

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$\left\{\begin{array}{ccc}y=-\frac{1}{3}x+6\\x=-6\end{array}\right\\\text{put the value of x to the first equation}\\\\y=-\dfrac{1}{3}\cdot(-6)+6=2+6=8\\\\\text{Answer:}\ x=-6;\ y=8\to(-6;\ 8)$

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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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3 years ago

Jeff drives 14 miles to school but his sister only drives 5 7/10 miles. How much farther does Jeff drive than his sister?

Anna007 [38]

14 - 5 7/10 = 13 10/10 - 5 7/10 = 8 3/10 <==

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3 years ago