What is a Parrel line?
Answer:
parrel line never meet
I need to find the equal expression to -m(2m+2n)+3mn+2m². Help please?
[tex]m(2m+2n)+3mn+2m^2\implies \stackrel{\textit{distributing}}{2m^2+2mn}+3mn+2m^2 \\\\\\ 2m^2+2m^2+2mn+3mn\implies \stackrel{\textit{adding like-terms}}{4m^2+5mn}[/tex]
Given the functions below, find (g•h) (1).
g(x) = х^2 +4+ 2х
h(x) = — 3х + 2
-7
-30
35
7
Answer:
-7
Step-by-step explanation:
We are given the following functions:
[tex]g(x) = x^2 + 4 + 2x[/tex]
[tex]h(x) = -3x + 2[/tex]
(g•h) (1)
The multiplication is:
[tex](g \times h)(1) = g(1) \times h(1)[/tex]
So
[tex]g(x) = 1^2 + 4 + 2(1) = 7[/tex]
[tex]h(1) = -3(1) + 2 = -3 + 2 = -1[/tex]
Then
[tex]g(1) \times h(1) = 7(-1) = -7[/tex]
So -7 is the answer.
If 21% of kindergarten children are afraid of monsters, how many out of
each 100 are afraid?
Answer:
The appropriate answer is "21".
Step-by-step explanation:
Given:
Afraid percentage,
p = 21%
or,
= 0.21
Sample size,
n = 100
As we know,
⇒ [tex]X=np[/tex]
By putting the values, we get
[tex]=0.21\times 100[/tex]
[tex]=21[/tex]
Consider the following functions. f(x) = x2, g(x) = x + 9 Find (f ∘ g)(x). Find the domain of (f ∘ g)(x). (Enter your answer using interval notation.) Find (g ∘ f)(x). Find the domain of (g ∘ f)(x). (Enter your answer using interval notation.) Find (f ∘ f)(x). Find the domain of (f ∘ f)(x). (Enter your answer using interval notation.) Find (g ∘ g)(x). Find the domain of (g ∘ g)(x). (Enter your answer using interval notat
Answer:
Whe we have two functions, f(x) and g(x), the composite function:
(f°g)(x)
is just the first function evaluated in the second one, or:
f( g(x))
And the domain of a function is the set of inputs that we can use as the variable x, we usually start by thinking that the domain is the set of all real numbers, unless there is a given value of x that causes problems, like a zero in the denominator, for example:
f(x) = 1/(x + 1)
where for x = -1 we have a zero in the denominator, then the domain is the set of all real numbers except x = -1.
Now, we have:
f(x) = x^2
g(x) = x + 9
then:
(f ∘ g)(x) = (x + 9)^2
And there is no value of x that causes problems here, so the domain is the set of all real numbers, that, in interval notation, is written as:
x ∈ (-∞, ∞)
(g ∘ f)(x)
this is g(f(x)) = (x^2) + 9 = x^2 + 9
And again, here we do not have any problem with a given value of x, so the domain is again the set of all real numbers:
x ∈ (-∞, ∞)
(f ∘ f)(x) = f(f(x)) = (f(x))^2 = (x^2)^2 = x^4
And for the domain, again, there is no value of x that causes a given problem, then the domain is the same as in the previous cases:
x ∈ (-∞, ∞)
(g ∘ g)(x) = g( g(x) ) = (g(x) + 9) = (x + 9) +9 = x + 18
And again, there are no values of x that cause a problem here, so the domain is:
x ∈ (-∞, ∞)
jos3ph has 16 meters of rope he wants to cut pieces of rope that are 0.2meters long how many prices can be cut
A 3.2
B8
C32
D80
Answer:
D.80
Step-by-step explanation:
You need to divide thus
16m/0.2m=80m
what is the sum factor of 3600
Answer:
Step-by-step explanation:
Answer:
24
Step-by-step explanation:
Find the prime factorization of the number 3,600. Factor Tree.
2|3,600.
2|1,800.
2|900.
2|450
5|225
5|45
3|9
3|3
|1
Setup the equation for determining the number of factors or divisors.
3600=2x2x2x2x3x3x5
Sum factors=2+2+2+2+2+3+3+5=24
I need the answer plzzzzzz
Answer:
answer is 0 in the ones column
1a. If an escape room party
has 16 participants and 4
escape puzzles:
• How many staff are
needed?
• Write an expression to
solve how many staff
are needed.
Answer:
2 staff members
Step-by-step explanation:
Given
See attachment for missing details
Let
[tex]s \to staff\ member[/tex]
[tex]p \to participant[/tex]
[tex]e \to puzzle[/tex]
Required
Staff members for 18 participants
From the attachment, we have:
[tex]1s \to 8p[/tex] ---- 1 staff member to 8 participants
[tex]s \to 8p[/tex]
Multiply both sides by 1
[tex]s * 2 \to 8p * 2[/tex]
[tex]2s \to 16p[/tex]
This means that 2 staff members are required for 16 participants
Annual windstorm losses, X and Y, in two different regions are independent, and each is uniformly distributed on the interval [0, 10]. Calculate the covariance of X and Y, given that X+ Y < 10.
Answer:
[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]
Step-by-step explanation:
Given
[tex]Interval =[0,10][/tex]
[tex]X + Y < 10[/tex]
Required
[tex]Cov(X,Y)[/tex]
First, we calculate the joint distribution of X and Y
Plot [tex]X + Y < 10[/tex]
So, the joint pdf is:
[tex]f(X,Y) = \frac{1}{Area}[/tex] --- i.e. the area of the shaded region
The shaded area is a triangle that has: height = 10; width = 10
So, we have:
[tex]f(X,Y) = \frac{1}{0.5 * 10 * 10}[/tex]
[tex]f(X,Y) = \frac{1}{50}[/tex]
[tex]Cov(X,Y)[/tex] is calculated as:
[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]
Calculate E(XY)
[tex]E(XY) =\int\limits^X_0 {\int\limits^Y_0 {\frac{XY}{50}} \, dY} \, dX[/tex]
[tex]X + Y < 10[/tex]
Make Y the subject
[tex]Y < 10 - X[/tex]
So, we have:
[tex]E(XY) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{XY}{50}} \, dY} \, dX[/tex]
Rewrite as:
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {XY}} \, dY} \, dX[/tex]
Integrate Y
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{XY^2}{2}}} }|\limits^{10 - X}_0 \, dX[/tex]
Expand
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2} - \frac{X(0)^2}{2}}} }\ dX[/tex]
[tex]E(XY) =\frac{1}{50}\int\limits^{10}_0 {\frac{X(10 - X)^2}{2}}} }\ dX[/tex]
Rewrite as:
[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X(10 - X)^2\ dX[/tex]
Expand
[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 X*(100 - 20X + X^2)\ dX[/tex]
[tex]E(XY) =\frac{1}{100}\int\limits^{10}_0 100X - 20X^2 + X^3\ dX[/tex]
Integrate
[tex]E(XY) =\frac{1}{100} [\frac{100X^2}{2} - \frac{20X^3}{3} + \frac{X^4}{4}]|\limits^{10}_0[/tex]
Expand
[tex]E(XY) =\frac{1}{100} ([\frac{100*10^2}{2} - \frac{20*10^3}{3} + \frac{10^4}{4}] - [\frac{100*0^2}{2} - \frac{20*0^3}{3} + \frac{0^4}{4}])[/tex]
[tex]E(XY) =\frac{1}{100} ([\frac{10000}{2} - \frac{20000}{3} + \frac{10000}{4}] - 0)[/tex]
[tex]E(XY) =\frac{1}{100} ([5000 - \frac{20000}{3} + 2500])[/tex]
[tex]E(XY) =50 - \frac{200}{3} + 25[/tex]
Take LCM
[tex]E(XY) = \frac{150-200+75}{3}[/tex]
[tex]E(XY) = \frac{25}{3}[/tex]
Calculate E(X)
[tex]E(X) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{X}{50}}} \, dY} \, dX[/tex]
Rewrite as:
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {X}} \, dY} \, dX[/tex]
Integrate Y
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 { (X*Y)|\limits^{10 - X}_0 \, dX[/tex]
Expand
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)] - [X * 0])\ dX[/tex]
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 ( [X*(10 - X)]\ dX[/tex]
[tex]E(X) =\frac{1}{50}\int\limits^{10}_0 10X - X^2\ dX[/tex]
Integrate
[tex]E(X) =\frac{1}{50}(5X^2 - \frac{1}{3}X^3)|\limits^{10}_0[/tex]
Expand
[tex]E(X) =\frac{1}{50}[(5*10^2 - \frac{1}{3}*10^3)-(5*0^2 - \frac{1}{3}*0^3)][/tex]
[tex]E(X) =\frac{1}{50}[5*100 - \frac{1}{3}*10^3][/tex]
[tex]E(X) =\frac{1}{50}[500 - \frac{1000}{3}][/tex]
[tex]E(X) = 10- \frac{20}{3}[/tex]
Take LCM
[tex]E(X) = \frac{30-20}{3}[/tex]
[tex]E(X) = \frac{10}{3}[/tex]
Calculate E(Y)
[tex]E(Y) =\int\limits^{10}_0 {\int\limits^{10 - X}_0 {\frac{Y}{50}}} \, dY} \, dX[/tex]
Rewrite as:
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 {\int\limits^{10 - X}_0 {Y}} \, dY} \, dX[/tex]
Integrate Y
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 { (\frac{Y^2}{2})|\limits^{10 - X}_0 \, dX[/tex]
Expand
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] - [\frac{(0)^2}{2}])\ dX[/tex]
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 ( [\frac{(10 - X)^2}{2}] )\ dX[/tex]
[tex]E(Y) =\frac{1}{50}\int\limits^{10}_0 [\frac{100 - 20X + X^2}{2}] \ dX[/tex]
Rewrite as:
[tex]E(Y) =\frac{1}{100}\int\limits^{10}_0 [100 - 20X + X^2] \ dX[/tex]
Integrate
[tex]E(Y) =\frac{1}{100}( [100X - 10X^2 + \frac{1}{3}X^3]|\limits^{10}_0)[/tex]
Expand
[tex]E(Y) =\frac{1}{100}( [100*10 - 10*10^2 + \frac{1}{3}*10^3] -[100*0 - 10*0^2 + \frac{1}{3}*0^3] )[/tex]
[tex]E(Y) =\frac{1}{100}[100*10 - 10*10^2 + \frac{1}{3}*10^3][/tex]
[tex]E(Y) =10 - 10 + \frac{1}{3}*10[/tex]
[tex]E(Y) =\frac{10}{3}[/tex]
Recall that:
[tex]Cov(X,Y) = E(XY) - E(X) \cdot E(Y)[/tex]
[tex]Cov(X,Y) = \frac{25}{3} - \frac{10}{3}*\frac{10}{3}[/tex]
[tex]Cov(X,Y) = \frac{25}{3} - \frac{100}{9}[/tex]
Take LCM
[tex]Cov(X,Y) = \frac{75- 100}{9}[/tex]
[tex]Cov(X,Y) = -\frac{ 25}{9}[/tex]
Am I correct if not plz asap help I have less Than 4 minutes
reflectiion across y=x
9514 1404 393
Answer:
see attached
Step-by-step explanation:
The reflection across y=-x swaps the coordinates and negates both of them. The first-quadrant figure becomes a third-quadrant figure.
(x, y) ⇒ (-y, -x)
A money box contains only 10-cent
and 20-cent coins. There are 28
coins with a total value of $3.80.
How many coins of each?
Answer:
Number of 10 cents = 18
Number of 20 cents = 10
Step-by-step explanation:
Let number of 10 cents be = x
Let number 20 cents be = y
Total number of coins = x + y = 28 -------- ( 1 )
Total amount in the box = 0.10 x + 0.20y = 3.80 ---------- ( 2 )
Solve the equations to find x and y
( 1 ) => x + y = 28
x = 28 - y
Substitute x in ( 2 )
( 2 ) => 0.10(28 - y) + 0.20y = 3.80
2.80 - 0.10y + 0.20y = 3.80
0.10 y = 3.80 - 2.80
0.10 y = 1.00
[tex]y = \frac{1}{0.10} = 10[/tex]
y = 10
Substitute y in ( 1 ) => x + y = 28
x + 10 = 28
x = 28 - 10
x = 18
Tom had some blocks that were all the same size and shape. He used two of them to make this regular hexagon He placed six more blocks around this hexagon to make a bigger regular hexagon
How many more blocks does he need to place around this shape to make the next bigger regular hexagon?
(A) 6
(B) 10
(C) 12
(D) 18
Answer:
Well it all started by drawing some equilateral triangles so that they made a regular hexagon: hexagon from unit length triangles. Then we ...
A bakers bread calls for 2 1/2 cups of flour he plans to make 2 loaves of bread.
measuring 1/3 cup how many scoops will he need
9514 1404 393
Answer:
15
Step-by-step explanation:
(2 loaves)(5/2 cups/loaf)/(1/3 cup/scoop) = (2)(5/2)(3/1) scoops = 15 scoops
The baker will need 15 scoops of flour.
A candy distributor needs to mix a 40% fat-content chocolate with a 60% fat-content chocolate to create 150 kilograms of a 52% fat-content chocolate. How many kilograms of each kind of chocolate must they use?
Answer:
60 kg of 40% and 90 kg of 60%
Step-by-step explanation:
Let the amount of 40% chocolate be x.
Let the amount of 60% chocolate be y.
x + y = 150
0.4x + 0.6y = 0.52 * 150
x + y = 150
4x + 6y = 780
-4x - 4y = -600
(+) 4x + 6y = 780
--------------------------
2y = 180
y = 90
x + y = 150
x + 90 = 150
x = 60
Answer: 60 kg of 40% and 90 kg of 60%
Compute the product AB by the definition of the product ofmatrices, where Ab1 and Ab2 are computed separately, and by therow-column rule for computing AB.
Matrix A= [2 -2]
[3 4]
[4 -3]
Matrix B =
[4 -1]
[-1 2]
Answer:
[tex]A * B = \left[\begin{array}{ccc}10&-6\\8&5\\19&-10\end{array}\right][/tex]
Step-by-step explanation:
Given
[tex]A =\left[\begin{array}{cc}2&-2\\3&4\\4&-3\end{array}\right][/tex]
[tex]B = \left[\begin{array}{cc}4&-1\\-1&2\end{array}\right][/tex]
Required
[tex]AB[/tex]
To do this, we simply multiply the rows of A by the column of B;
So, we have:
[tex]A * B = \left[\begin{array}{ccc}2*4 + -2*-1&2*-1+-2*2\\3*4+4*-1&3*-1+4*2\\4*4-3*-1&4*-1-3*2\end{array}\right][/tex]
[tex]A * B = \left[\begin{array}{ccc}10&-6\\8&5\\19&-10\end{array}\right][/tex]
2x-5y=22n y=3x-7 Use substitution to solve the system.
Answer:
x = 1 , y = -4
Step-by-step explanation:
2x - 5y = 22 ------- ( 1 )
y = 3x - 7 ------- ( 2 )
Substitute ( 2 ) in ( 1 ) :
2x - 5 (3x - 7) = 22
2x - 15x + 35 = 22
- 13x = 22 - 35
- 13x = - 13
x = 1
Substitute x in ( 1 ) :
2x - 5y = 22
2 ( 1 ) - 5y = 22
- 5y = 22 - 2
-5y = 20
y = - 4
The box plots show the weights, in pounds, of the dogs in two different animal shelters.
Weights of Dogs in Shelter A
2 box plots. The number line goes from 6 to 30. For the weights of dogs in shelter A, the whiskers range from 8 to 30, and the box ranges from 17 to 28. A line divides the box at 21. For shelter B, the whiskers range from 10 to 28, and the box ranges from 16 to 20. A line divides the box at 18.
Weights of Dogs in Shelter B
Which is true of the data in the box plots? Select three choices.
The median weight for shelter A is greater than that for shelter B.
The median weight for shelter B is greater than that for shelter A.
The data for shelter A are a symmetric data set.
The data for shelter B are a symmetric data set.
The interquartile range of shelter A is greater than the interquartile range of shelter B.
Answer:
The median weight for shelter A is greater than that for shelter B.
The data for shelter B are a symmetric data set.
The interquartile range of shelter A is greater than the interquartile range of shelter B.
Step-by-step explanation:
The median weight for shelter A is greater than that for shelter B.
The median of A = 21 and the median of B = 18 true
The median weight for shelter B is greater than that for shelter A.
The median of A = 21 and the median of B = 18 false
The data for shelter A are a symmetric data set.
False, looking at the box it is not symmetric
The data for shelter B are a symmetric data set.
true, looking at the box it is symmetric
The interquartile range of shelter A is greater than the interquartile range of shelter B.
IQR = 28 - 17 = 11 for A
IQR for B = 20 -16 = 4 True
Fourteen children out of a group of 26 like chocolate ice cream. What would be the numerator of the fraction illustrating proportion of children in this group that do not
like chocolate ice cream?
Answer:
12
Step-by-step explanation:
The amount of children that do like ice cream are 14/26 so the children that do not like ice cream 14/26, and the numerator is 12
The length of a rectangle should be 9 meters longer than 7 times the width. If the length must be
between 93 and 163 meters long, what are the restrictions for the width, p?
Write the solution set as an algebraic inequality solved for the variable.
Answer:
If we define W as the width:
12m ≤ W ≤ 22m
Step-by-step explanation:
We have a rectangle with length L and width W.
We know that:
"The length of a rectangle should be 9 meters longer than 7 times the width"
Then:
L = 9m + 7*W
We also know that the length must be between 93 and 163 meters long, so:
93m ≤ L ≤ 163m
Now we want to find the restrictions for the width W.
We start with:
93m ≤ L ≤ 163m
Now we know that L = 9m + 7*W, then we can replace that in the above inequality:
93m ≤ 9m + 7*W ≤ 163m
Now we need to isolate W.
First, we can subtract 9m in the 3 sides of the inequality
93m - 9m ≤ 9m + 7*W -9m ≤ 163m -9m
84m ≤ 7*W ≤ 154m
Now we can divide by 7 in the 3 sides, so we get:
84m/7 ≤ 7*W/7 ≤ 154m/7
12m ≤ W ≤ 22m
Then we can conclude that the width is between 12 and 22 meters long.
WILL GIVE BRAINLIEST
15 POINTS
Determine how the triangles can be proven similar.
AA~
SSS~
SAS~
Not similar
AA~ as both the triangles are congruent because of vertically opposite angles and alternative interior angles .
Simplify this algebraic expression.
Z-4/4+8
O A. Z+7
O B. z+ 9
O c. z-3
O D. Z-7
Answer:
Z +7
Step-by-step explanation:
Z-4/4+8
Divide first
Z -1 +8
Add and subtract
Z +7
Given the functions:
g(n) = 3n - 5
f(n) = n2 + 50
Find:
(g+f)(8)
Answer:
[tex](g + f)(8) =133[/tex]
Step-by-step explanation:
Given
[tex]g(n) = 3n - 5[/tex]
[tex]f(n) = n^2 + 50[/tex]
Required
[tex](g + f)(8)[/tex]
This is calculated as:
[tex](g + f)(n) =g(n) + f(n)[/tex]
So, we have:
[tex](g + f)(n) =3n - 5 + n^2 +50[/tex]
[tex]Substitute[/tex] 8 for n
[tex](g + f)(8) =3*8 - 5 + 8^2 +50[/tex]
[tex](g + f)(8) =24 - 5 + 64 +50[/tex]
[tex](g + f)(8) =133[/tex]
Which best describes the range of the function f(x) = 2(3)x?
Answer:
y > 0.
Step-by-step explanation:
A.
Pedro and his friend Cody played basketball in the backyard. Cody made 5 Baskets . Pedro made 15 baskets. How many times more baskets did pedro make than cody?
Answer: 10
Step-by-step explanation: 15 - 5 = 10
Suppose the mean percentage in Algebra 2B is 70% and the standard deviation is 8% What percentage of students receive between a 70% and 94% enter the value of the percentage without the percent sign
Answer:
49.87
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Suppose the mean percentage in Algebra 2B is 70% and the standard deviation is 8%.
This means that [tex]\mu = 70, \sigma = 8[/tex]
What percentage of students receive between a 70% and 94%
The proportion is the p-value of Z when X = 94 subtracted by the p-value of Z when X = 70. So
X = 94
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{94 - 70}{8}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a p-value of 0.9987.
X = 70
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{70 - 70}{8}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a p-value of 0.5.
0.9987 - 0.5 = 0.4987.
0.4987*100% = 49.87%.
So the percentage is 49.87%, and the answer, without the percent sign, is 49.87.
You measure 37 dogs' weights, and find they have a mean weight of 69 ounces. Assume the population standard deviation is 9.2 ounces. Based on this, construct a 90% confidence interval for the true population mean dog weight.Give your answers as decimals, to two places_______ ± ________ ounces (Enter your answer as an integer or decimal number. Examples: 3, -4, 5.5172 ; Enter DNE for Does Not Exist, oo for Infinity)
Solution :
Given :
mean weight of the dogs, [tex]$\overline x$[/tex] = 69
Number of dogs, n = 37
Standard deviation, σ = 9.2
Confidence interval = 90%
At 90% confidence interval for the true population mean dog weight is given by :
[tex]$= \overline x \pm \frac{\sigma}{\sqrt n} \times z_{0.05}$[/tex]
[tex]$= 69\pm \frac{9.2}{\sqrt_{37}}} \times 1.64485$[/tex]
[tex]$=69 \pm 2.487799$[/tex]
= (66.5122, 71.4878)
As part of a board game, players choose 5 unique symbols from 9 different symbols to create their secret password. How many different ways can the players create a specific 5 symbol password?
Give your answer in simplest form.
Answer:
[tex]15,120[/tex]
Step-by-step explanation:
For the first symbol, there are 9 options to choose from. Then 8, then 7, and so on. Since each player chooses 5 symbols, they will have a total of [tex]9\cdot 8 \cdot 7 \cdot 6\cdot 5=\boxed{15,120}[/tex] permutations possible. Since the order of which they choose them matters (as a different order would be a completely different password), it's unnecessary to divide by the number of ways you can rearrange 5 distinct symbols. Therefore, the desired answer is 15,120.
Answer:15,120
Step-by-step explanation:
Somebody please help asap
Answer:
B. [tex] 4x^2 + \frac{3}{2}x - 7 [/tex]
Step-by-step explanation:
[tex] f(x) = \frac{x}{2} - 3 [/tex]
[tex] g(x) = 4x^2 + x - 4 [/tex]
(f + g)(x) = f(x) - g(x)
= [tex] \frac{x}{2} - 3 + 4x^2 + x - 4 [/tex]
Add like terms
[tex] = 4x^2 + \frac{x}{2} + x - 3 - 4 [/tex]
[tex] = 4x^2 + \frac{3x}{2} - 7 [/tex]
[tex] = 4x^2 + \frac{3}{2}x - 7 [/tex]
