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1 year ago

Find the first partial derivative: f(x, y) = y^5 -3xy

1 answer:

8 0

Partial derivative is derivative over only 1 variable.
here we have 2 variables so it depends on which variable you want to make partial derivative.

case 1.
df/dx = -3y

case 2
df/dy = 5y^4 - 3x

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Ben Carson was loaned $125.00 to pay $6.00 interest for $100.00 every week ,after 3 weeks how much do he pay back to clear the d

Ghella [55]

Answer:

Total amount pay after 3 week = $147.5

Step-by-step explanation:

Given:

Amount of loan taken = $125

Amount interest per $100 = $6 per week

Find:

Total amount pay after 3 week

Computation:

Amount of interest per week = 125[6/100]

Amount of interest per week = $7.5 per week

Amount of interest in three week = 3 x Amount of interest per week

Amount of interest in three week = 3 x 7.5

Amount of interest in three week = $22.5

Total amount pay after 3 week = Amount of loan taken + Amount of interest in three week

Total amount pay after 3 week = 125 + 22.5

Total amount pay after 3 week = $147.5

8 0

1 year ago

In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.

WARRIOR [948]

Answer:

For First Solution: $y_1(t)=e^t$

$y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution: $y_2(t)=cosht$

$y_2(t)=cosht$ is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: $y_1(t)=e^t$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_1(t)=e^t$

First order derivative:

$y'_1(t)=e^t$

2nd order Derivative:

$y''_1(t)=e^t$

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence $y_1(t)=e^t$ is the solution of equation y''-y=0.

For 2nd Solution:

$y_2(t)=cosht$

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

$y_2(t)=cosht$

First order derivative:

$y'_2(t)=sinht$

2nd order Derivative:

$y''_2(t)=cosht$

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence $y_2(t)=cosht$ is the solution of equation y''-y=0.

3 0

1 year ago

**WARNING** IF YOU ANSWER THIS QUESTION INCORRECTLY OR TRY TO ANSWER WITH FOOLISHNESS YOU WILL GET REPORTED

frozen [14]

The solution is attached as a word file.

Download doc

3 0

1 year ago

E varies directly with the square root of C. If E=40 when C=25, find: C when E = 10.4

sukhopar [10]

Answer: C = 1.69

Step-by-step explanation:

E is proportional to √C

To remove proportionality, introduce a constant (k).

E = k × √C

From question,

E = 40 and C = 25

So,

40 = k ×√25

40 = k × 5

k = 8

Now,

C = ?

E = 10.4

k = 8

E = k × √C

10.4 = 8 × √C

10.4 / 8 = √C

( 10.4 / 8 ) ^ 2 = C

C = 1.69

6 0

2 years ago

Read 2 more answers

Which graph represents the function f(x)=x4+4x3−12x2−32x+64?

sp2606 [1]

$f(x)=x^4+4x^3-12x^2-32x+64$

The degree of f(x) is 4. Also the leading coefficient is 1 and it is positive

So as x approaches infinity then y approaches infinity

as x approaches -infinity then y approaches infinity

The first and fourth graph goes up and it satisfies the above . so we ignore the second and third graph.

Now we check the x intercepts of the first graph

x intercepts of first graph is -4 and 2

Plug in -4 for x in f(x) and check whether we get 0

$f(x)=x^4+4x^3-12x^2-32x+64$

$f(x)=(-4)^4+4(-4)^3-12(-4)^2-32(-4)+64=0$

Now plug in 2 for x and check

$f(x)=(2)^4+4(2)^3-12(2)^2-32(2)+64=0$

So -4 and 2 are the x intercepts that satisfies f(x)

Hence first option is the graph of $f(x)=x^4+4x^3-12x^2-32x+64$

5 0

1 year ago

Read 2 more answers