All categories

3 years ago

Need help combing like terms

Mathematics

1 answer:

ki77a [65]3 years ago

8 0

8ab + ab - 2b -b

Let me know if you need it in simplest form!

You might be interested in

Determine the equation of the line that has an x intercept of 5 and y intercept of 7

garik1379 [7]

Answer:

y = −1.4x + 7

Step-by-step explanation:

There are 3 steps to find the Equation of the Straight Line

1. Find the slope of the line

2. Put the slope and one point into the "Point-Slope Formula"

3. Simplify

3 0

2 years ago

Suppose you have the following recursion formula a1 = 1, a2 = 2, and an = a(n - 1)+ a(n - 2) for integers n ≥ 3. How would you d

Firlakuza [10]

$\left\{\begin{array}{ccc}a_1=1\\a_2=2\\a_n=a_{n-1}+a_{n-2}&for\ n\geq3\end{array}\right\\\\a_3=a_{3-1}+a_{3-2}=a_2+a_1\to a_3=2+1=3\\\\a_4=a_{4-1}+a_{4-2}=a_3+a_2\to a_4=3+2=5\\\\a_5=a_{5-1}+a_{5-2}=a_4+a_3\to a_5=5+3=8\\\vdots$

6 0

3 years ago

if layla can right 20 essays in 15 minutes and 8 in 6 minutes then how many can she right in 45 minutes?

scoray [572]

Answer:

60! hope this helps!

Step-by-step explanation:

4 0

3 years ago

Please answer fast however answer i will make brainilest

Marizza181 [45]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

A tank has the shape of a surface generated by revolving the parabolic segment y = x2 for 0 ≤ x ≤ 3 about the y-axis (measuremen

Darina [25.2K]

Answer:

$100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy$ ft-lbs.

Step-by-step explanation:

Given:

The shape of the tank is obtained by revolving $y=x^2$ about y axis in the interval $0\leq x\leq 3$ .

Density of the fluid in the tank, $D=100\ lbs/ft^3$

Let the initial height of the fluid be 'y' feet from the bottom.

The bottom of the tank is, $y(0)=0^2=0$

Now, the height has to be raised to a height 5 feet above the top of the tank.

The height of top of the tank is obtained by plugging in $x=3$ in the parabolic equation . This gives,

$H=3^2=9\ ft$

So, the height of top of tank is, $y(3)=H=9\ ft$

Now, 5 ft above 'H' means $H+5=9+5=14$

Therefore, the increase in height of the top surface of the fluid in the tank is given as:

$\Delta y=(14-y)$ ft

Now, area of cross section of the tank is given as:

$A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section$

Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.

We have, $y=x^2$

So, $x=\sqrt y$

Therefore, radius, $r=\sqrt y$

Now, area of cross section is, $A(y)=\pi (\sqrt y)^2$

Work done in pumping the contents to 5 feet above is given as:

$W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy$

Plug in all the values. This gives,

$W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}$

7 0

3 years ago