Solve using any method
Write the equation of the line which passes through (1, 2) and with slope of - 1 3 in standard form.
I have already completed the question but I am confused on how I got the the answer. y−y1 = m(x−x1) y−2 = -1 3 •(x−1) y−2 = -1 3 x+ 1 3 1 3 x+y...
I don't understand the question. What would the equation be for this vertex of the parabola? I'm doing Conic sections, and I don't understand the question.
I also need to know the end-behavior. for example- x----> + or - infinity
Can you help me with this question? The equation for the line that passes through (-3,5) and (-2,-6)
I cannot remember how to factor to the cubic term
A solution is 10% citric acid. if 3 liters of a 20% citric acid solution are added to it, the result is a solution this is in a ratio of 13 parts citric acid to 102 parts "other than citric acid"...
i don't know this answer can you help me?
select the approxiate value of x that is the solution to f(x)=0 where f(x)=-9x^2+6x+7
select the approxiate value of x that is solutions for (x)=0 value f(x)=-3x^2+2x+8
I know you could plug everything in and figure it out that way but I can't do that because I don't have another point on the parabola. For example: vertex:(-1,-3) focus:(-1,0)
Select he value of x that are solutions to the inequality 0>x^2+5x-2 1. xE (-5-33 squared/2,-5+33 squared/2) 2. xE [-5-33 squared/2,-5+33...
Discuss/Explain how the graph F(x) = -2f(x+1) -3 can be obtained from the graph f(x). If (0,5), (6,7), and (-9,-4) are on the graph of f. where do they end up on the graph of F?
simplify write answer in form a+bi, where a and b are real numbers
(a) Given 6^3=216, write an equivalent statement in logarithmic form. (b) Given log_2?√8=3/2, write an equivalent statement in exponential form
1. The circle below is centered at the origin, and the length of its radius is 3. What is the equation of the circle in standard form?
I have been having trouble with Algebra 2 problems like this.
Find the present value of $3,000.00 if interest is 6% compounded quarterly for 5 years