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Vinvika [58]
3 years ago
5

Calculate the following derivatives. Assume y is a function of t . Use Y for the derivative of y . (a) d dt y7= Incorrect: Your

answer is incorrect. (b) d dt y3e4t= Incorrect: Your answer is incorrect. (c) d dt t7 cos(y6)=
Mathematics
1 answer:
xxTIMURxx [149]3 years ago
7 0

Answer:

(a) d dt y7=7Y^6

(b) d dt y3e4t=3Y^2e^{4t}+4Y^3e^{4t}

(c) d dt t7 cos(y6)=7t^6 cos(y6)+t^7(-sin(y^6))·6Y^5

Step-by-step explanation:

We calculate the following derivatives. Since these are complex derivatives, we also calculate the formula for complex derivatives. We use  Y for the derivative of y . Therefore, we calculate and we get

(a) d dt y7=7Y^6

(b) d dt y3e4t=3Y^2e^{4t}+4Y^3e^{4t}

(c) d dt t7 cos(y6)=7t^6 cos(y6)+t^7(-sin(y^6))·6Y^5

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Solve using the quadratic formula: Y = 2x^2 - 7x-3
algol13

Answer:

x=-0.39 x=3.89

Step-by-step explanation:

a=2

b=-7

c=-3

when you plug all of those numbers in you would get

(7±\sqrt{73} )/ (4)

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Tom is afraid of heights above 9 feet. He is asked to repair a side of a high deck. The bottom of a ladder must be placed 6 feet
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3 years ago
Let F = (2, 3). Find coordinates for three points that are equidistant from F and the y-axis. Write an equation that says P = (x
True [87]

Answer:

The equation that says P is equidistant from F and the y-axis is P(x,y) =\left(1,\frac{3+y'}{2} \right).

(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.

Step-by-step explanation:

Let F(x,y) = (2,3) and R(x,y) =(0, y'), where P(x,y) is a point that is equidistant from F and the y-axis. The following vectorial expression must be satisfied to get the location of that point:

F(x,y)-P(x,y) = P(x,y)-R(x,y)

2\cdot P(x,y) = F(x,y)+R(x,y)

P(x,y) = \frac{1}{2}\cdot F(x,y)+\frac{1}{2} \cdot R(x,y) (1)

If we know that F(x,y) = (2,3) and R(x,y) = (0,y'), then the resulting vectorial equation is:

P(x,y) = \left(1,\frac{3}{2} \right)+\left(0, \frac{y'}{2}\right)

P(x,y) =\left(1,\frac{3+y'}{2} \right)

The equation that says P is equidistant from F and the y-axis is P(x,y) =\left(1,\frac{3+y'}{2} \right).

If we know that y_{1}' = -3, y_{2}' = 0 and y_{3}' = 3, then the coordinates for three points that are equidistant from F and the y-axis:

P_{1}(x,y) = \left(1,\frac{3+y_{1}'}{2} \right)

P_{1}(x,y) = (1,0)

P_{2}(x,y) = \left(1,\frac{3+y_{2}'}{2} \right)

P_{2}(x,y) = \left(1,\frac{3}{2} \right)

P_{3}(x,y) = \left(1,\frac{3+y_{3}'}{2} \right)

P_{3}(x,y) = \left(1,6 \right)

(1, 0), (1, 3/2) and (1,6) are three points that are equdistant from F and the y-axis.

7 0
3 years ago
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