11/04/20
The quotient of a number x and 3 is at most 4.
An inequality. Write the word sentence as an inequality. Then solve the inequality.
11/04/20
Linear Algebra, advanced mathematics
Okay, let e1 = (1,0), e2 = (0,1), y1 = (2,5), and y2=(-1,6). Let T: R2 →R2 be a transformation that maps e1 into y1 and maps e2 into y2. Find the image of (5, -3) under T. Is T onto? Is it...
more
11/03/20
Why does the equation y = 3x + 5 NOT represent a proportional relationship? 7th grade math
This is 7th grade math.
11/03/20
How do I apply the distributive property to factor out the greatest common factor for 25+50
11/03/20
Is 7xy a group in the set R?
Hey,So i think i'm getting to grips with this but just need to check my logic is correct here.Is the following a group, The set of all real number R, with the binary operation 7xy.It appears to...
more
11/03/20
What is the quotient of ⅔ and ½ ?
11/03/20
Function: ℎ(𝑛)=3𝑛−1
11/03/20
When you add 18 to 1/4 of a number you get the number itself
When you add 18 to 1/4 of a number you get the number
11/02/20
What is the correct answer to the following questions?
Let A = {−3, −2, −1, 0, 1, 2, 3, 4, 5, 6} and define a relation R on A as follows:For all x, y A, x R y ⇔ 3|(x − y).It is a fact that R is an equivalence relation on A. Use set-roster notation to...
more
11/02/20
4(3x+2) distributive proporty
11/02/20
the sum of 6 and one-fifth of x
10/31/20
Show x+y/xy is not a group
οHi,I need help to show that this is not a group(positive real numbers, ο) where xοy=(x+y)/xyI know it's closed as x and y are positive real numbers, so not equal to 0, so xy cannot be 0, therefore...
more
10/31/20
Vector Spaces - Linear Algebra
Let R and C be the field of real numbers and the field of complex numbers, respectively. Let Mm,n(R) be the set of all m × n matrices over R and let Mn(R) = Mn,n(R). Let R 3 = {(x, y, z) : x, y, z...
more
Still looking for help? Get the right answer, fast.
Ask a question for free
Get a free answer to a quick problem.
Most questions answered within 4 hours.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.