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forsale [732]
3 years ago
8

The graph shows the rate at which the depth of the water in a pond is changing over time. On a coordinate plane, a graph titled

Depth of Pond Water has Minutes on the x-axis and Feet on the y-axis. A line goes through points (2, 4) and (4, 8). The depth of the water is increasing by feet each minute.
Mathematics
1 answer:
lawyer [7]3 years ago
4 0

Answer:

Step-by-step explanation:

The key to solving this question is to find the distance bwtween the two points given

distance=y2-y1/x2-x1

(2,4)(4,8)

2=x1,4=y1

4=x2

8=y2

8-4/4-2

4/2=2

The depth of the water is increasing by 2 ft each minute

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Answer:

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Step-by-step explanation:

1 min 12 seconds is 72 seconds. 28 meters on 72 seconds. 28×2.5=72

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Find gradient <br><br>xe^y + 4 ln y = x² at (1, 1)​
cricket20 [7]

xe^y+4\ln y=x^2

Differentiate both sides with respect to <em>x</em>, assuming <em>y</em> = <em>y</em>(<em>x</em>).

\dfrac{\mathrm d(xe^y+4\ln y)}{\mathrm dx}=\dfrac{\mathrm d(x^2)}{\mathrm dx}

\dfrac{\mathrm d(xe^y)}{\mathrm dx}+\dfrac{\mathrm d(4\ln y)}{\mathrm dx}=2x

\dfrac{\mathrm d(x)}{\mathrm dx}e^y+x\dfrac{\mathrm d(e^y)}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

e^y+xe^y\dfrac{\mathrm dy}{\mathrm dx}+\dfrac4y\dfrac{\mathrm dy}{\mathrm dx}=2x

Solve for d<em>y</em>/d<em>x</em> :

e^y+\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x

\left(xe^y+\dfrac4y\right)\dfrac{\mathrm dy}{\mathrm dx}=2x-e^y

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x-e^y}{xe^y+\frac4y}

If <em>y</em> ≠ 0, we can write

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2xy-ye^y}{xye^y+4}

At the point (1, 1), the derivative is

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Evaluate cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.
Andre45 [30]

Answer:

Option d)  5 to the power of negative 5 over 6 is correct.

\dfrac{\sqrt[3]{\bf 5} \times \sqrt{\bf 5}}{\sqrt[3]{\bf 5^{\bf 5}}}= 5^{\frac{\bf -5}{\bf 6}}

Above equation can be written as 5 to the power of negative 5 over 6.

ie, 5^\frac{\bf -5}{\bf 6}

Step-by-step explanation:

Given that cube root of 5 multiplied by square root of 5 over cube root of 5 to the power of 5.

It can be written as below

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}} \times 5^{\frac{1}{2}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{1}{3}+\frac{1}{2}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= \dfrac{5^{\frac{2+3}{6}}}{5^{\frac{5}{3}}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5}{6}} \times 5^{\frac{-5}{3}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{\sqrt[3]{5^5}}= 5^{\frac{5-10}{6}}

\dfrac{\sqrt[3]{5} \times \sqrt{5}}{5^5}= 5^{\frac{-5}{6}}

Above equation can be written as 5 to the power of negative 5 over 6.

7 0
4 years ago
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