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2 years ago

D is directly proportional to t2 When t = 4, D = 8 Find the positive value of t when D = 50

Mathematics

2 answers:

alekssr [168]2 years ago

5 0

Answer: either 46 or 25 is your answer

Step-by-step explanation: Brainliest pls

olasank [31]2 years ago

3 0

Answer:

t = 10

Step-by-step explanation:

Proportions

D is directly proportional to t squared. This can be written as:

$D = kt^2$

Where k is the proportionality constant. It can be calculated by substituting the given condition in the formula, when t=4, D=8:

$8 = k4^2=16k$

Solving for k:

$k=\frac{8}{16}=\frac{1}{2}$

Thus, the proportional relationship is:

$\displaystyle D = \frac{1}{2}t^2$

When D=50, the value of t can be found by solving for t:

$\displaystyle 50 = \frac{1}{2}t^2$

Multiplying by 2:

$t^2=100$

Taking the square root:

$t=\sqrt{100}$

The positive value of t is:

t = 10

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2 years ago

If 2y=x³ + 2x², find dy/dx when x=2.

scZoUnD [109]

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Step-by-step explanation:

$\boxed{\textsf{If }y=x^n,\: \textsf{then }\dfrac{\text{d}y}{\text{d}x}=nx^{n-1}}$

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$2y=x^3+2x^2$

Isolate y by dividing both sides by 2:

$\implies y=\dfrac{1}{2}x^3+x^2$

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$\implies \dfrac{\text{d}y}{\text{d}x}=3 \cdot \dfrac{1}{2}x^{3-1}+2 \cdot x^{2-1}$

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Finally, substitute x = 2 into the differentiated equation:

$\begin{aligned} \implies \dfrac{\text{d}y}{\text{d}x} & =\dfrac{3}{2}(2)^2+2(2)\\\\ & = \dfrac{3}{2}(4)+4\\\\ & = \dfrac{12}{2}+4\\\\ & = 6+4\\\\ & = 10\end{aligned}$

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1 year ago

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:

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

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

3 0

2 years ago