The answer is -1 because if you see the graph
Answer:
The solutions to the system of equations are:
Step-by-step explanation:
Given the system of the equations
Isolate x for for 4x-3y=-13
Divide both sides by 4
substitute y=5
The solutions to the system of equations are:
Answer:
A) Ed
Explanation: How i got it was by first looking at what Eric’s weight loss program, I then guessed he lost about the same amount of weight each week (1.5), i then did that times 4 since that’s how many weeks were recorded on the chart which gave me 6. I then looked at the chart again and saw that Ed first started with the weight 205 and by week 4 his weight was 197, I then did 205-197 which equals 8, I then decided that Ed lost more weight.
Answer:
x-intercept: (18/5,0)
y-intercept: (0,-2)
Step-by-step explanation:
btw .. you put the equation twice it's supposed to be " -5x+9y=-18 "
Answer:
The integrals was calculated.
Step-by-step explanation:
We calculate integrals, and we get:
1) ∫ x^4 ln(x) dx=\frac{x^5 · ln(x)}{5} - \frac{x^5}{25}
2) ∫ arcsin(y) dy= y arcsin(y)+\sqrt{1-y²}
3) ∫ e^{-θ} cos(3θ) dθ = \frac{e^{-θ} ( 3sin(3θ)-cos(3θ) )}{10}
4) \int\limits^1_0 {x^3 · \sqrt{4+x^2} } \, dx = \frac{x²(x²+4)^{3/2}}{5} - \frac{8(x²+4)^{3/2}}{15} = \frac{64}{15} - \frac{5^{3/2}}{3}
5) \int\limits^{π/8}_0 {cos^4 (2x) } \, dx =\frac{sin(8x} + 8sin(4x)+24x}{6}=
=\frac{3π+8}{64}
6) ∫ sin^3 (x) dx = \frac{cos^3 (x)}{3} - cos x
7) ∫ sec^4 (x) tan^3 (x) dx = \frac{tan^6(x)}{6} + \frac{tan^4(x)}{4}
8) ∫ tan^5 (x) sec(x) dx = \frac{sec^5 (x)}{5} -\frac{2sec^3 (x)}{3}+ sec x
