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3 years ago

Find the limit. lim t→0 5et − 5 t , 1 + t − 1 t , 4 1 + t

Mathematics

1 answer:

sergeinik [125]3 years ago

5 0

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A 15-foot ladder is placed against a vertical wall of a building, with the bottom of the ladder standing on level ground 14 feet

Finger [1]

Answer:

Step-by-step explanation:

Well certainly friction alone will not hold it there.

c² = a² + b²

15² = 14² + b²

15² - 14² = b²

b² = 29

b = √29 = 5.385164... = 5.39 ft

6 0

3 years ago

Tye is making some trail mix Based on a recipe that combines 2 cups of nuts with 3 cups of granola. If Tye Has 6 cups of nuts, h

WINSTONCH [101]

Answer:

9 cups

Step-by-step explanation:

Ratio of cups of nuts to cups of granola = 2:3

This means that total ratio = 2+3 = 5

Proportion of nuts = 2/5

Proportion of granola = 3/5

So, if Tye has 6 cups of nuts, then 6 represent a proportion of 2/5

then 3/5 would represent how many cups? = $\frac{3}{5} *6 * \frac{5}{2}\\ \\$

The 5 at the top and 5 at the bottom cancel out and you are left with;

=(3*6) /2

=18 /2

= 9 cups

4 0

3 years ago

A car is traveling at 112 km/h when the driver sees an accident 70 m ahead and slams on the brakes. What minimum constant decele

ANTONII [103]

velocity=distance/time

31.1m/s=70/t

t=70/31.1

t=2.25secs

acceleration=velocity/time

a=31.1/2.25

a=13.82ms-²

therefore the constance deceleration=13.82ms-²

3 0

3 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$