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3 years ago

Which of the following is a sequence that moves on a basis of addition and subtraction?

Mathematics

1 answer:

irinina [24]3 years ago

5 0

It is the first answer

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Express this number in scientific notation 763 thousandths

yan [13]

It would be 763 x 10^-4

3 0

3 years ago

Find the first partial derivatives of the function f(x,y,z)=4xsin(y−z)

Amanda [17]

Answer:

$f_x(x,y,z)=4\sin (y-z)$

$f_x(x,y,z)=4x\cos (y-z)$

$f_z(x,y,z)=-4x\cos (y-z)$

Step-by-step explanation:

The given function is

$f(x,y,z)=4x\sin (y-z)$

We need to find first partial derivatives of the function.

Differentiate partially w.r.t. x and y, z are constants.

$f_x(x,y,z)=4(1)\sin (y-z)$

$f_x(x,y,z)=4\sin (y-z)$

Differentiate partially w.r.t. y and x, z are constants.

$f_y(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial y}(y-z)$

$f_y(x,y,z)=4x\cos (y-z)$

Differentiate partially w.r.t. z and x, y are constants.

$f_z(x,y,z)=4x\cos (y-z)\dfrac{\partial}{\partial z}(y-z)$

$f_z(x,y,z)=4x\cos (y-z)(-1)$

$f_z(x,y,z)=-4x\cos (y-z)$

Therefore, the first partial derivatives of the function are $f_x(x,y,z)=4\sin (y-z), f_x(x,y,z)=4x\cos (y-z)\text{ and }f_z(x,y,z)=-4x\cos (y-z)$ .

4 0

3 years ago

A scientist launched an object upward at a speed of 32 ft per second from a height of 240 feet. The height of the object is mode

QveST [7]

Answer:

B 5 seconds

Step-by-step explanation:

h = 240 + 32t – 16t^2

When is this equation 0

0 = 240 + 32t – 16t^2

Factor out 16

0 = -16(-15-2t+t^2)

= -16(t^2-2t-15)

What 2 numbers multiply to -15 and add to -2

-5*3 = -15

-5+3 =-2

0 = -16(t-5) (t+3)

Using the zero product property

t-5=0 t+3=0

t=5 t=-3

Since time cannot be negative

t=5 seconds

3 0

3 years ago

The diameter of a cylindrical water tank is 8ft and its height is 11ft what is the volume of the tank?

Sedbober [7]

Answer:

2211.68123

Step-by-step explanation:

Using the formula to find volume of cylinder

V=πr2h=π·82·11≈2211.68123

3 0

2 years ago

Jack is mowing a lawn that has a shed. needs to know area of lawn to mow. dimensions of yard are 5x by 13 x-4 . she'd are 2x by

Elena-2011 [213]

The diagram of the lawn and the shed is shown below.

The area of the lawn needed to be mowed equals to the area of the yard minus the area of the shed

The area of the yard = $(5x)(13x-4) = 65 x^{2} -20x$
The area of the shed = $(2x)(2x+4)=4 x^{2} +8x$

The area of the lawn = $65 x^{2} -20x-(4 x^{2} +8x)$
The area of the lawn = $65 x^{2} -20x-4 x^{2} -8x$
The area of the lawn = $61 x^{2} -28x$

5 0

2 years ago