Harley begins solve this problem by combining like terms 3x + 5x =10

Find the perimeter of the trapezoid

Olin [163]

Answer:

1st slot: 12

2nd slot: 48

Step-by-step explanation:

Hope this helps!

8 0

2 years ago

Find the domain of the function.<br><br><br><br> f(x) = -5x + 3

Sedaia [141]

Enter a problem...

Calculus Examples

Popular Problems Calculus Find the Domain and Range f(x)=5x-3

f

(

x

)

=

5

x

−

3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

(

−

∞

,

∞

)

Set-Builder Notation:

{

x

|

x

∈

R

}

The range is the set of all valid

y

values. Use the graph to find the range.

Interval Notation:

(

−

∞

,

∞

)

Set-Builder Notation:

{

y

|

y

∈

R

}

Determine the domain and range.

Domain:

(

−

∞

,

∞

)

,

{

x

|

x

∈

R

}

Range:

(

−

∞

,

∞

)

,

{

y

|

y

∈

R

}

3 0

2 years ago

If P(-4,-2) and Q(1, -3) represent points in a coordinate

Artyom0805 [142]

Answer:

the answer is (6, -4)

Step-by-step explanation:

4 0

3 years ago

What is the slope of the graph?

Dennis_Churaev [7]

Answer:

-4

Step-by-step explanation:

Slope can be found with $\frac{y2-y1}{x2-x1}$

Pick two points on the graph that the line runs through

(-2, 4) will fill the values of x1 and y1, and (-1, 0) will fill the values of x2 and y2

Insert the coordinates into the formula

$\frac{0-4}{(-1)-(-2)}$

$\frac{-4}{-1+2}$

$\frac{-4}{1}$

Therefore the slope is -4

3 0

2 years ago

Water is leaking out of an inverted conical tank at a rate of 12,000 cm3/min at the same time that water is being pumped into th

Bas_tet [7]

Answer:

$291111.1cm^3/min$

Step-by-step explanation:

We are given that

$\frac{dV}{dt}_{out}=12000 cm^3/min$

Height of tank,h=6 m

Diameter of top,d=4 m

Radius,r= $\frac{d}{2}=\frac{4}{2}=2 m$

$\frac{dh}{dt}=20 cm/min$

$\frac{r}{h}=\frac{2}{6}=\frac{1}{3}$

$r=\frac{1}{3} h$

We have to find rate at which water is being pumped into the tank.

Volume of conical ,V= $\frac{1}{3}\pi r^2 h$

$V=\frac{1}{3}\pi(\frac{1}{3} h)^2h=\frac{1}{27}\pi h^3$

$\frac{dV}{dt}=\frac{1}{9}\pi h^2(\frac{dh}{dt})$

h=2 m=200 cm

1m=100 cm

$\frac{dV}{dt}_{in}-12000=\frac{1}{9}\pi(200)^2\times 20$

$\frac{dV}{dt}_{in}=12000+\frac{1}{9}\pi (40000)\times 20$

$\frac{dV}{dt}_{in}=291111.1cm^3/min$

4 0

3 years ago