2 Algebra Questions Check My Answer And Help On The Other One

2 Algebra Questions!! please check my answer and help on the other one!!!?

1) A No solution
The reason that A is correct is because the other points listed above do not work in either equation(as you said) and you need a point that works in both equations. If both points would have worked in both equations, then the answer would have been D.

2) A (3, -½)
4x+2y=11
x-2=-2y <--- simply divide this equation by -2
-½x+1=y <--- Plug this in whenever you see y(lets use the first equation)
--------------
4x+2(-½x+1)=11 <--- Simply multiply
4x-x+2=11 <--- Subtract 2 from both sides and combine like terms of x
3x=9 <---divide by 3
x=3 <---x term; Plug this back into one of the original equations to find the y value(Let's use the second one)
---------
(3)-2=-2y <---3-2=1
1=-2y <---Divide by -2
-½=y <---y term

So your answer is A, (3,-½)

I hope I have helped you and you can now do problems like these with more ease. Have a happy Friday! :)

Can some one check my algebra answers for these two questions?

1)Find the domain and range of:
a) g(x)=I2x-3I-3
b) h(x)=3x+2/x-1
c) f(x)=x^4-4
Answer
a) domain=all real numbers
range= all real numbers
b) domain=(negative infinity,1)u(1, positive infinity)
range=(negative infinity,3)u(3,positive infinity)
c) domain=all real numbers
range=(-4,positive infinity)

2) If the endpoints of a diameter of a circle are (4,3)and (-3,5), what is the equation of the circle?

Answer
(5-3)^2+(-3-4)^2
=2^2+(-7)^2
=4+49
=53

radius=53/2
midpoint : (1/2, 4)
equation of the circle=(x-1/2)^2+(y-4)^2=53/4

Algebra 2 help please! Check my answers?!?

I don't know if I did these right or not so can you check my answers for me and if I didn't tell me the right answer and how to get it please and thank you.

1.Write the expression in simplified radical form. Show your work.
3-2√11/2+√11

My Answer = 3


2.Solve the equation. Show your check then write the solution set.
x-1=√6x+10

My answer = x-1 = sqrt(6) sqrt(x)+10

3.Give the complex number, 5 – 3i:
a)Graph the complex number in the complex plane
b)Calculate the modulus. When necessary, round to the tenths place.

My Answer = I don't know how to do this one...

Algebra II questions?

1. Rewrite it, part of it got cut off

2.
(x^2)+(2y^2)=33
x^2/33 + 2y^2/33 = 1
x^2/33 + y^2/(33/2) = 1

Thus you have an ellipse with x intercepts (solutions) at sqrt(33) and -sqrt(33).

3. 5|3x-4|=x+1
|3x - 4| = (x + 1)/5

3x - 4 = (x + 1)/5 and 3x - 4 = -(x + 1)/5

15x - 20 = x + 1
14x = 21
x = 21/14 = 3/2 (answer to first equation)

3x - 4 = -(x + 1)/5
15x - 20 = -x - 1
16x = 19
x = 19/16

x = 3/2 and 19/16

4. ([3]/[5d])/([-9]/[15df])
(3/5d) * (15df/-9)
45df/-45d
-f

Check My Algebra Problems?

1. Find slope and y-intercepts
f(x)=-2x-10
the slope is ? -5
the y-intercept is? (0,-10)

2. System of Equations
x+3y=7 (1)
x=8-3y (2)
my answer (6,1/3)

3.Translate to an algebraic expression
The product of 31% and some number
My answer: 31%y

4. Solve by substitution method
2x+5y=11
x=33-8y
my answer: (-7,5)

I really need help with some algebra questions. Some I need to have checked, while others I need help with?

C(x) = 15+ .25x<<< My Answer. I agree.


Inverse<<Distributive -- this answer looks right.
Why?
4(2x − y + 1) − 3(4x − 2y − 1)
To solve the equation, you distribute the 4* to the 2x and -y and +1, for example.


Starting with C = 2πr, the problem wants to say what π is in terms of C and r. To do this, isolate the π term.
C = 2πr ==> C / 2r = 2πr / 2r ==> C / 2r = π


−5 < − 4x + 3 ≤ 23
−5 -3 < − 4x + 3 -3 ≤ 23 -3
−8 < − 4x ≤ 20
−8 / -4 < − 4x / -4 ≤ 20 / -4
2 > x >= -5
The first image fits this.


20 + 1.5g ≤ d<<

You can plug the numbers into the equations and see which equation is true for both points.
Luckily I tried the second one first, and it works. You should probably try the others, and see that both points /don't/ work for them. (Maybe there are /two/ right answers, who knows???)
2x + y = 2
trying (2, -2) 2 (2) + (-2) =? 2 .. 4 - 2 = 2
trying (-5, 12) 2 ( -5) + 12 =? 2 .. -10 + 12 = 2


C(x) = 480x + 1600<<< My Answer -- Right.


You have the function as this: (8x5y2+12x3y20x2y7)/(4xy2) but from the answers, I think you skipped some things. It should be this: (8x5y2+12x3y6+20x2y7)/(4xy2)
(8x5y2+12x3y5+20x2y7)/(4xy2)
The carat ^ is an accepted way of showing "to the power of", just so you know.
So,
(8x^5y^2+12x^3y^5+20x^2y^7)/(4xy^2)
= 8x^5y^2 / (4xy^2) + 12x^3y^5/ (4xy^2) + 20x^2y^7 / (4xy^2)
= 2 x^4 + 3 x^2 y^3 + 5 xy^5


You have to do this one in small chunks, I think.
A rectangle with the dimensions x by x plus three is inside a larger rectangle with the dimensions three x by x squared plus seven x plus twelve.
First, list the two areas.
Area 1 = x * (x+3)
Area 2 = 3x * (x^2 + 7x + 12)

ratio of area of small to large is: area 1 / area 2

So, the answer will be a reduced form of

x*(x+3) / [3x * (x^2 + 7x + 12) ]

The answer would be this one
(x+3)/(x2+7x+12)
except that I think there should still be a 3 in the denominator.

But maybe the problem wasn't rewritten just right? Anyway, that answer is closest to correct, from the info I see.

ALGEBRA 2. PLEASE HELP ME VERIFY THIS! :)?

First thank all of u who helped me on my previous question I was able to do almost all my problems after that except for one

-[2z-(5z+2)=2+(2z+7)

This is what I did but I don't know if its right please help :)

-[2z-(5z+2)]=2+(2z+7)
-(2z-5z-2)=2+2z+7
2z-5z-2z=2+7-2

z=7

Please let me know if that's right I have an exam next week your help is SUPER greatly appreciated :) THANK YOUUU :)