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Lana71 [14]
3 years ago
8

G use part 1 of the fundamental theorem of calculus to find the derivative of the function. sqrt(4+7t)

Mathematics
1 answer:
uysha [10]3 years ago
6 0
Let y = √4+7t
then u= 4+7t
y=√u = u^½

du/dt= 7
dy/du = ½U^-½

dy/dt = du/dt • dy/du
= 7×½U^-½
= 7/2√U
= 7 / (2{√4+7t})
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Answer:

The answer is

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y =  - 4

Step-by-step explanation:

We must find a solution where

\frac{6}{y + 1}  +  \frac{y}{y - 2}  =  \frac{6}{y + 1}  \times  \frac{y}{y - 2}

Consider the Left Side:

First, to add fraction multiply each fraction on the left by it corresponding denomiator and we should get

\frac{6}{y + 1}  \times  \frac{y - 2}{y - 2}  +  \frac{y}{y - 2}  \times  \frac{y + 1}{y + 1}

Which equals

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Add the fractions

\frac{y {}^{2} + 7y - 12 }{(y - 2)(y + 1)}  =  \frac{6}{y + 1}  \times  \frac{y}{y - 2}

Simplify the right side by multiplying the fraction

\frac{6y}{(y  + 1)(y + 2)}

Set both fractions equal to each other

\frac{6y}{(y + 1)(y - 2)}  =  \frac{ {y}^{2} + 7y - 12}{(y + 1)(y - 2)}

Since the denomiator are equal, we must set the numerator equal to each other

6y =  {y}^{2}  + 7y - 12

=  {y}^{2}  + y - 12

(y  + 4)(y - 3)

y =  - 4

y = 3

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3 years ago
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Answer:

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Step-by-step explanation:

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