Here is the best answerrrrr.
out of every 13 will most likely be an ace.
Answer:
The answer to your question is y = 4/3x + 1/3
Step-by-step explanation:
Data
Point A = (2, 3)
Point B = (5, 7)
Process
1.- Calculate the slope
x1 = 2 y1 = 3
x2 = 5 y2 = 7
m = (y2 - y1)/(x2 - x1)
- Substitution
m = (7 - 3)/(5 - 2)
- Slope
m = 4/3
2.- Find the equation of the line
y - y1 = m(x - x1)
y - 3 = 4/3(x - 2)
y - 3 = 4/3x - 8/3
y = 4/3x - 8/3 + 3
y = 4/3x - 8/3 + 9/3
y = 4/3x + 1/3
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
Answer: $187 will be in the account after 6 years.
Step-by-step explanation:
We would apply the formula for determining compound interest which is expressed as
A = P(1+r/n)^nt
Where
A = total amount in the account at the end of t years
r represents the interest rate.
n represents the periodic interval at which it was compounded.
P represents the principal or initial amount deposited
From the information given,
P = $100
r = 11% = 11/100 = 0.11
n = 1 because it was compounded once in a year.
t = 6 years
Therefore,.
A = 100(1 + 0.11/1)^1 × 6
A = 100(1 + 0.11)^6
A = 100(1.11)^6
A = $187
