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LenaWriter [7]
3 years ago
9

Alexis and Tasha challenged each other to a

Mathematics
2 answers:
Anvisha [2.4K]3 years ago
8 0

Answer:

45 WPM

Step-by-step explanation:

A = 1.2T

54 = 1.2T

T = 45

Marat540 [252]3 years ago
8 0

Answer: 45 WPM

Step-by-step explanation: 54 dived by 1.2 = 45

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option b

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Question 3. Y=(1/5)^xSketch the graph of each of the exponential functions and label three points on each graph.
STatiana [176]

Given exponential function:

y\text{ = (}\frac{1}{5})^x

Let us obtain three points including the y-intercept so that we can plot the function y = f(x)

When x =0:

\begin{gathered} y\text{ = (}\frac{1}{5})^0 \\ =\text{ 1} \end{gathered}

when x =1:

\begin{gathered} y\text{ = (}\frac{1}{5})^1 \\ =\text{ }\frac{1}{5} \end{gathered}

when x =2:

\begin{gathered} y\text{ = (}\frac{1}{5})^2 \\ =\text{ }\frac{1}{25} \end{gathered}

We have the points : (0, 1), (1, 1/5), and (2, 1/25)

Using these points, let us provide a sketch of the plot of y =f(x). We have the plot as shown below:

5 0
1 year ago
John and Pam are paid 38.25 for their work. John worked 1.5 hours and Pam worked 3 hours. They split the money according to the
Nina [5.8K]

John's share of the money is 12.75.

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7 0
3 years ago
Read 2 more answers
How to differentiate ?
Bas_tet [7]

Use the power, product, and chain rules:

y = x^2 (3x-1)^3

• product rule

\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}

• power rule for the first term, and power/chain rules for the second term:

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}

• power rule

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3

Now simplify.

\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) =  \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)

Differentiate both sides and you end up with the same derivative:

\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2

7 0
2 years ago
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