Which expression is the greatest common factor (GCF) of the terms of trinomial 12x^7y^9 + 6x^4y^7 - 10x^3y^5
Answers
Answer: 2. [tex]\mathbf{2x^3y^5}[/tex]
Step-by-step explanation:
Greatest common factor of any two or more expression is the largest common expression that divides them.The given expression : [tex]12x^7y^9 + 6x^4y^7 - 10x^3y^5[/tex]
Term 1 : [tex]12x^7y^9[/tex]
Term 2 : [tex]6x^4y^7[/tex]
Term 3: [tex]- 10x^3y^5[/tex]
Lowest power of x = 3
Thus , The highest common power of x = [tex]x^3[/tex] (i)
Lowest power of y = 5
The highest common power of y =[tex]y^5[/tex] (ii)
Greatest common factor of 12, 6 , -10= 2 [Because 2 is the largest number that divides all of them] (iii)
From (i) , (ii) , (iii) , we have
Greatest common expression : [tex]2x^3y^5[/tex]
Hence, the correct answer is 2. [tex]\mathbf{2x^3y^5}[/tex]
Related Questions
If 52/x is a positive integer, how many integer values are possible for x?
Answers
There are a total of 6 integer values that are possible for x: 1, 2, 4, 13, 26, and 52.
What is the quotient?If 52/x is a positive integer, it means that 52 is divisible by x, and x is a factor of 52. In other words, x must be a positive divisor of 52.
The positive divisors of 52 are: 1, 2, 4, 13, 26, and 52.
Thus;
52/1 = 52 (a positive integer)
52/2 = 26 (a positive integer)
52/4 = 13 (a positive integer)
52/13 = 4 (a positive integer)
52/26 = 2 (a positive integer)
52/52 = 1 (a positive integer)
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The probability that you will win a game is 0.18. if you play the game 504 times, what is the most likely number of wins?
Answers
What's 10⁄12 written as a fraction in simplest form?
A. 5⁄6
B. 10⁄6
C. 3⁄5
D. 5⁄12
Answers
A. 5/6
Which question is not a good survey question?
a) Don’t you agree that the financial crisis is essentially over?
b) On average, how many hours do you sleep per day?
c) What is your opinion of educational funding this year?
d) Are you happy with the availability of electronic products in your state?
Answers
Answer:
See image
Step-by-step explanation:
Plato
what is e/4 + 2f-3 when e =12 andf =1/2
Answers
When putting in a number for a variable, we want to put it on parenthesis. So, let's write the problem with the variables substituted.
[tex](12)/4+2(1/2)-3[/tex]
We could also write it like this:
[tex]\frac{\left(12\right)}{4}+2\left(\frac{1}{2}\right)-3=1[/tex]
To start, we have to remove the parenthesis.
[tex]=\frac{12}{4}+2\cdot \frac{1}{2}-3[/tex]
Let's simplify the first fraction.
[tex]\frac{12}{4}=3[/tex]
Let's simplify the second part:
[tex]2\cdot \frac{1}{2}[/tex]
[tex]=\frac{1\cdot \:2}{2}[/tex]
[tex]=\frac{2}{2}[/tex]
[tex]=1[/tex]
We would be left with,
[tex]=3+1-3[/tex]
Simplify.
[tex]=1[/tex]
Thanks!
-TetraFish
Let's take a look at our question.
"What is e/4 + 2f -3 when e = 12 and f = 1/2?"
To solve for this, we need to plug in our values to their proper values.
Let's replace "e" with "12" and "f" with "0.5" (0.5 is equal to 1/2).
Our expression should look like this now:
12/4 + 2(0.5) -3
Well, we can simplify two parts of our expression: 12/4 and 2(0.5).
We can simplify 12/4 into 3, as 12 divided by 4 is 3.
We can also simplify 2(0.5) into 1, as multiplying by 0.5 is the same as dividing by 2.
We now have:
3 + 1 - 3
We can cancel out 3 and -3 to make 0.
We are now left with 1.
Your answer remains as "1".
I hope this helps! :)
M(5, 7) is the midpoint of rs The coordinates of S are (6, 9). What are the coordinates of R?
(5.5, 8)
(7, 11)
(10, 14)
(4, 5)
Answers
[tex]x_M = \frac{x_R + x_S}{2}[/tex]
[tex]5 = \frac{x + 6}{2}[/tex]
[tex]10 = x + 6[/tex]
[tex]x = 4[/tex]
[tex]y_M = \frac{y_R + y_S}{2}[/tex]
[tex]7 = \frac{y + 9}{2}[/tex]
[tex]14 = y + 9[/tex]
[tex]y = 5[/tex]
[tex]\therefore R(4, 5)[/tex]
Which pair of non-congruent figures must be similar? two squares two parallelograms (not rectangles) two right triangles two isosceles triangles (not equilateral)
Answers
Find the percent equivalent to the ratio 8 to 20
Answers
12 to 25 ==> 12/25 = 0.48 ===> 0.48*100 = 48%
divide the ratio:
8/20 = 0.4
0.4 = 40%
Using 8(-20) in a real world situation
Answers
Let (-7, -4) be a point on the terminal side of theta. Find the exact values of Sin, CSC and COT
Answers
To get the radius, we us the following formula.
r = √(-7)² + (-4)²
This reads:
radius = square root of -7 to the power of 2 plus -4 to the power of 2.
I am new here and I am not sure if they support latex. Other wise I would of used latex.
Ok so r (radius) = r = √(-7)² + (-4)² = √65
r = √65
Now we have our radius we can find the values for sin csc and cot
sin = y / r
csc = r / y
cot = x / y
I will do the first one for you which is sin.
sin = y / r = -4 / √65
The above is saying sin = y divided by r = -4 divided by the square root of 65.
If you need help with the others let me know.
If 4 1/2 = 2, 8 1/3 = 2, and 16 1/4 = 2, then for what value of x would x 1/5 = 2?
2
4
24
32
Answers
By paying attention to the hole number and ignoring the fraction. You can see the pattern is doubling each time. 4, 8, 16, 32.
Find an ordered pair (x , y) that is a solution to the equation
Answers
if x = 0, y = 3
ordered pair (0,3)
The amount of carbon-14 present in animal bones t years after the animal's death is given by P(t)= -0.00012097t. How old is an ivory tusk that has lost 34% of its carbon-14?
Answers
I believe the equation you gave is wrong because the standard form of equation for C-14 decay is in the form of:
A = Ao e^-kt
So I think the right form of equation is (correct me if I’m wrong):
P(t) = Po e^(-0.00012097t)
Where,
Po = initial value of C-14 at t = 0
t = time elapsed
Since it is given that:
P = (1 – 0.34) Po
P = 0.66 Po
Therefore, t is:
0.66 Po = Po e^(-0.00012097 t)
0.66 = e^(-0.00012097 t)
taking ln of both side:
ln 0. 66 = -0.00012097 t
t = - ln 0. 66 / 0.00012097
t = 3,434.86 years
Therefore the ivory tusk is about 3,435 years old.
If, on average, Bob can make a sale to every 3rd person that comes into his store, how many people must come into Bob�s store if he wanted to make approximately fifteen (15) sales?
Answers
n = 3 * 15 = 45
"let v = r 2 with the usual addition and scalar multiplication defined by k(u1, u2) = (ku1, 0). determine which of the five axioms of vector spaces involving scalar multiplication v satisfies and which fail. for the ones it satisfies, prove that it satisfies the axiom. for those that fail, show that it fails with a counterexample."
Answers
[tex]\mathbf u=(u_1,u_2)[/tex]
and let [tex]k\in\mathbb R[/tex] be any real constant.
Given this definition of scalar multiplication, we can see right away that there is no identity element [tex]e[/tex] such that
[tex]e\mathbf u=\mathbf u[/tex]
because
[tex]e\mathbf u=e(u_1,u_2)=(eu_1,0)\neq(u_1,u_2)=\mathbf u[/tex]
a,(-a)^2=a^2 This is true or false?
Answers
What is the median of the data set given below? 19, 22, 46, 24, 37, 16, 19, 33
Answers
16,19,19,22,24,33,37,46....the median is the middle number. If there is an even number of numbers, like this one, there will be 2 middle numbers. Find those 2 middle numbers, divide them by 2, and that is ur median.
start from the ends and count inwards till u come to the 2 middle numbers...
(22 + 24) / 2 = 46/2 = 23
** now if there would have been an odd number of numbers, u would just have 1 middle number, and that number would be the median.
16,19,19, 22, 24, 33, 37, 46
If the total amount of data [let's take it as n] is odd, then the median is data on [[tex] \frac{n}{2} [/tex]].
But here the total amount of data is even [n=8], so the median is the average of the data on [[tex] \frac{n}{2} [/tex]] and [[tex] \frac{n}{2} +1[/tex]].
So the median is
[tex] \frac{22+24}{2} = \frac{46}{2} = 23[/tex]
An ellipse has vertices along the major axis at (0, 8) and (0, –2). The foci of the ellipse are located at (0, 7) and (0, –1). What are the values of a, b, h, and k, given the equation below? (y-k)^2/a^2+(x+h)^2/b^2=1
Answers
now hmm, from the provided vertices and focus point, you can pretty much see what "a" is, half of the major axis, is just 5.
now, the center is from either vertex to half-way up, or "a" units up, so say from -2 + 5, is at 3, so the center is at 0, 3.
now, the distance from a focus point to the center, is 4 units, like say from 0, 3 up to 0,7.
[tex]\bf \textit{ellipse, vertical major axis}\\\\ \cfrac{(y-{{ h}})^2}{{{ a}}^2}+\cfrac{(x-{{ k}})^2}{{{ b}}^2}=1 \qquad \begin{cases} center\ ({{ h}},{{ k}})\\ vertices\ ({{ h}}, {{ k}}\pm a)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{{{ a }}^2-{{ b }}^2}\\ ----------\\ h=0\\ k=3\\ a=5\\c=4 \end{cases} \\\\\\ \cfrac{(y-3)^2}{5^2}+\cfrac{(x-0)^2}{b^2}=1[/tex]
now, let' s find "b".
[tex]\bf c=\sqrt{a^2-b^2}\implies c^2=a^2-b^2\implies b^2=a^2-c^2 \\\\\\ b=\sqrt{a^2-c^2}\implies b=\sqrt{5^2-4^2}\implies b=3[/tex]
so, just plug that in.
Answer: The values for a, b, h, and k are a = 5, b = 3, h = 0, k = -3.
Step-by-step explanation: In this problem, we know ellipse has vertices along the major axis at (0, 8) and (0, -2). The foci of the ellipse are located at (0, 7) and (0, -1). We are asked to determine the values of a, b, h, and k.
We were also then provided with the equation for vertical eclipse:
[tex]\frac{(x-h)^2}{b^2} + \frac{(y -k)^2}{a^2}[/tex]
Before we begin, we need to first define our values for a, b, h, and k.
a - distance to vertices from the centerb - distance to co-vertices from the center(h, k) - represents the center of the eclipseThe first step, we need to determine the center of the eclipse. We can use the midpoint formula to determine the midpoint between the vertices along the major axis: (0, 8) and (0, -2).
[tex]M = (\frac{x_{1} +x_{2} }{2} , \frac{y_{1} + y_{2} }{2} )[/tex]
[tex]M = (\frac{0 + 0}{2} , \frac{-2 + 8}{2} )\\M = (0, 3)[/tex]
We now know that our center (h, k) is (0, 3). Which means our values for h and k are 0 and 3. Next, we have to determine our values for a and b. Considering the center of our eclipse is not at the center, we can use one of our vertices to determine our value for a.
[tex]V_{1}[/tex] = (h, k±a)
(0, 8) = (0, 3±a)
3 ± a = 8
±a = 5
Now, we know that a = 5. For us to get b, we need to use this formula: [tex]c^2 = a^2 - b^2[/tex]. Let's rewrite this formula, so we can focus on getting our b-value.
[tex]c^2 - a^2 = -b^2[/tex]
For us to use this formula, we need to determine our c value. To find our c-value, we have use of our foci points: (h, k±c). C is the units away/further from the center towards our foci points.
(0, 3±c) = (0, 7)
3 + c = 7
7 - 3 = c
4 = c
Now, we know that our value for c is 4. Now, let's plug into the formula.
[tex](4)^2 - (5)^2 = -b^2\\16 - 25 = -b^2\\\frac{-9}{-1} = \frac{-b^2}{-1} \\\sqrt{b^2} = \sqrt{9} \\b = 3[/tex]
Our value for b is 3. If we put into our eclipse formula:
[tex]\frac{(x-0)^2}{3^2} + \frac{(y -(-3))^2}{5^2}[/tex]
Line segment AB is congruent to line segment CD.
A.AB overbar similar to CD overbar
B.AB overbar congruent to CD overbar
C. AB overbar equal to CD overbar
D. AB overbar element to CD overbar
Answers
When two line segments are congruent, what it really means is that the length of the line segments are equal. Congruent is usually interchanged with the word equal since it means the same: equal lengths. However, congruent is very specific in the sense that it means “equal lengths”. When we say equal alone, it may refer to different properties, congruent however is very specific.
Now the answer to this question should be:
B. AB overbar congruent to CD overbar
Not similar, not equal, and definitely not element. What we mean in this case is “equal lengths”, therefore the perfect word is congruent.
How many different committees can be formed from 10 teachers and 41 students if the committee consists of 2 teachers and 2 students?
Answers
where C(n,r) represents the number of ways (order not important) we can choose r objects out of n, then
Number of ways to choose teachers = 10 choose 2 = C(10,2), and
number of ways to choose students = 41 choose 2 = C(41,2)
So the number of different committees
= C(10,2)*C(41,2)
= 45*820
= 36900
what is the answer to 11 over 7 in improper fractions
Answers
Factor of x^3-x^2-24x-36
Answers
If we let f(x) = x³ - x² - 24x - 36 and all we have to do is sub' in values of x until
f(x) = 0, we can use this to find an initial factor by the factor theorem:
f(1) = (1)³ - (1)² - 24(1) - 36 = -60
f(2) = (2)³ - (2)² - 24(2) - 36 = -80
f(5) = (5)³ - (5)² - 24(5) - 36 = -56
*** f(6) = (6)³ - (6)² - 24(6) - 36 = 0 ***
f(6) = 0 so (x - 6) is a factor of f(x).
This means that: f(x) = x³ - x² - 24x - 36 = (x - 6)(ax² + bx + c).
To find a,b and c, use long division (or inspection) to divide x³ - x² - 24x - 36 by x - 6.
The other 2 factors of f(x) can then be found by factorizing the
ax² + bx + c quadratic the way you would with any other quadratic (i.e. by quadratic formula, CTS or inspection).
Jane is one of 50 students to take a standardized math test that includes 100 multiple choice questions. If she has the highest score of any student with a raw score of 87, what is her percentile score?
Answers
Solution:
87 correct questions over 100 = 87/100 = .87 = (change to percent by moving the decimal 2 places to the right) 87%
The perimeter of a rectangle is 56 feet. Describe the the possible lengths of a side if the area of the rectangle is not to exceed 132 square ft
Answers
56 = 2(L + W)
56/2 = L + W
28 = L + W.....L = 28 - W
A = L * W
132 = W(28 - W)
132 = -W^2 + 28W
W^2 - 28W + 132 = 0
(W - 6)(W - 22) = 0
W - 6 = 0 L = 28 - W
W = 6 L = 28 - 6
L = 22
W - 22 = 0 L = 28 - W
W = 22 L = 28 - 22
L = 6
not sure which (length or width)....one of them is 6 ft and the other is 22 ft
The possible lengths of sides of the rectangle will be [tex](6,22)[/tex] or [tex](22,6)[/tex] .
What is area ?Area is the space which is occupied by any [tex]2D[/tex] surface.
Area of rectangle [tex]=Length\ *\ Breadth[/tex]
Perimeter of rectangle [tex]=2(Length\ +\ Breadth)[/tex]
We have,
Perimeter of rectangle [tex]=56[/tex] feet
Area of rectangle [tex]=132[/tex] squared feet
Let,
Length of rectangle [tex]=x[/tex]
Breadth of rectangle [tex]=y[/tex]
So,
Perimeter of rectangle [tex]=2(Length\ +\ Breadth)[/tex]
[tex]56=2(x+y)[/tex]
[tex]28=(x+y)[/tex]
⇒ [tex]x=28-y[/tex] [tex]........(i)[/tex]
Now,
Area of rectangle [tex]=Length\ *\ Breadth[/tex]
[tex]132=x\ *\ y[/tex] [tex]........(ii)[/tex]
Now substitute value of [tex]x[/tex] in [tex](ii)[/tex],
[tex]132=(28-y)y[/tex]
[tex]132=28y-y^2[/tex]
⇒[tex]y^2-28y+132=0[/tex]
Now using Middle term split method;
[tex]y^2-28y+132=0[/tex]
[tex]y^2-22y-6y+132=0[/tex]
[tex]y(y-22)-6(y-22)=0[/tex]
[tex](y-6)(y-22)=0[/tex]
⇒
[tex](y-6)=0[/tex]
⇒ [tex]y=6[/tex]
[tex](y-22)=0[/tex]
⇒ [tex]y=22[/tex]
So,
Now if we take [tex]y=6[/tex], then
[tex]x=28-6[/tex]
[tex]x=22[/tex]
And,
if we take [tex]y=22[/tex], then
[tex]x=28-22[/tex]
[tex]x=6[/tex]
So, the possible lengths of sides of the rectangle are [tex](6,22)[/tex] or [tex](22,6)[/tex] .
Hence, we can say that he possible lengths of sides of the rectangle are [tex](6,22)[/tex] or [tex](22,6)[/tex] .
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Solve the equation by graphing. x^2+14x+45=0 First, graph the associated parabola by plotting the vertex and four additional points, two on each side of the vertex. Then, use the graph to give the solution(s) to the equation. If there is more than one solution, separate them with commas.
Points:
Solution(s):
Answers
y=ax²+bx+c
the x value of the vertex is -b/(2a)
so
y=1x²+14x+45
vertex is -14/(2*1)=-14/2=-7
sub back to find y value
y=(-7)²+14(-7)+45
y=49-98+45
y=-4
the vertex is (-7,-4)
some other points are
(1,60)
(-1,32)
(-8,-3)
(-9,0)
we see that the solutions to y=0 and y=x^2+14x+45 is at x=-9 and x=-5
A total of 279 tickets were sold for the school play. they were either adult or student tickets. the number of student tickets sold was two times the number of adult tickets sold. how many adult tickets were sold?
Answers
a+s=279, using a from above in this equation gives you:
a+2a=279 combine like terms on left side
3a=279 divide both sides by 3
a=93
So 93 adult tickets were sold.
Final answer:
The number of adult tickets sold for the school play is 93, calculated by setting up an equation based on the information that the total tickets sold were 279 and student tickets were two times the number of adult tickets.
Explanation:
The question asks us to determine the number of adult tickets sold when a total of 279 tickets were sold for the school play, and the number of student tickets is two times the number of adult tickets sold. We can set up an equation to solve this problem.
Let A be the number of adult tickets, and S be the number of student tickets. The problem gives us two equations:
S = 2A (The number of student tickets is two times the number of adult tickets)
A + S = 279 (The total number of tickets sold is 279)
Substituting the first equation into the second gives us:
A + 2A = 279
3A = 279
A = 279 / 3
A = 93
Therefore, 93 adult tickets were sold for the school play.
A psychologist claims that less than 5.8 percent of the population suffers from professional problems due to extreme shyness. Express the null hypothesis and the alternative hypothesis in symbolic form. Use p, the true percentage of the population that suffers from extreme shyness. Select one: a. H0: p = 5.8% H1: p < 5.8% b. H0: p > 5.8% H1: p ≤ 5.8% c. H0: p < 5.8% H1: p ≥ 5.8% d. H0: p = 5.8% H1: p > 5.8%
Answers
Answer:
Step-by-step explanation:
Find the length of the third Angelou a triangle given that the first two angels are 35 and 70 show your work
Answers
angles in a triangle = 180 degrees
70 +35 = 105 degrees
180-105 = 75 degrees
3rd angle = 75 degrees
Which number is greater? 0 -1 -25 -50?
Answers
Remember that with positive numbers, 50 is greater than 1, but with negative numbers, -1 is greater than -50.
If that confuses you, think of it like this.
If you owe someone 1 dollar, you yourself have more than if you owe someone 50 dollars. You yourself have even more if you don't owe any money.
Guess the value of the limit (correct to six decimal places). (if an answer does not exist, enter dne.) lim hâ0 (4 + h)5 â 1024 h
Answers
Recall the definition of the derivative of a function [tex]f(x)[/tex] at a point [tex]x=c[/tex]:
[tex]f'(c)=\displaystyle\lim_{h\to 0}\frac{f(c+h)-f(c)}h[/tex]
We can then see that [tex]f(c)=c^5[/tex], and by the power rule we have [tex]f'(c)=5c^4[/tex]. Then replacing [tex]c=4[/tex], we arrive at
[tex]\displaystyle\lim_{h\to0}\frac{(4+h)^5-4^5}h=5\times4^4=1280[/tex]
Alternatively, we could have expanded the binomial, giving
[tex]\dfrac{(4+h)^5-4^5}h=\dfrac{(4^5+5\times4^4h+10\times4^3h^2+10\times4^2h^3+5\times4h^4+h^5)-4^5}h[/tex]
[tex]=\dfrac{1280h+640h^2+160h^3+20h^4+h^5}h[/tex]
[tex]=1280+640h+160h^2+20h^3+h^4[/tex]
and so as [tex]h\to0[/tex] we're left with 1280, as expected.