Answer:
The y intercept would be -4.
Plot as (0, - 4)
Answer:
zh
Step-by-step explanation:
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:cost of each pound of apple= $3
And cost of each pound of orange =$2
Step-by-step explanation:
Step 1
Let cost of apples = x
And cost of Oranges =y
Let 6 pounds of apples and 3 pounds of oranges cost 24 dollars be represented as
6 x + 3y= 24----- equation 1
Also, Let 5 pounds of apples and 4 pounds of oranges cost 23 dollars be represented as
5x+ 4y= 23----- equation 2
Step 2
6 x + 3y= 24----- equation 1
5x+ 4y= 23----- equation 2
Using substitution method to solve the equation
6 x + 3y= 24
24-6x=3y
y= 24-6x/3 = 8-2x
Y= 8-2x
Substituting the value of y= 8-2x into equation 2
5x+ 4( 8-2x)= 23
5x+ 32 -8x= 23
32-23= 8x-5x
9=3x
x=9/3
x=3
Putting the value of x= 3 in equation 1 and solving to find y
6 x + 3y= 24
6(3) +3y= 24
18+3y=24
3y= 24-18
3y=6
y=6/3= 2
Therefore the cost of each pound of apple= $3
And cost of each pound of orange =$2
Answer:
512
Step-by-step explanation:
Suppose we ask how many subsets of {1,2,3,4,5} add up to a number ≥8. The crucial idea is that we partition the set into two parts; these two parts are called complements of each other. Obviously, the sum of the two parts must add up to 15. Exactly one of those parts is therefore ≥8. There must be at least one such part, because of the pigeonhole principle (specifically, two 7's are sufficient only to add up to 14). And if one part has sum ≥8, the other part—its complement—must have sum ≤15−8=7
.
For instance, if I divide the set into parts {1,2,4}
and {3,5}, the first part adds up to 7, and its complement adds up to 8
.
Once one makes that observation, the rest of the proof is straightforward. There are 25=32
different subsets of this set (including itself and the empty set). For each one, either its sum, or its complement's sum (but not both), must be ≥8. Since exactly half of the subsets have sum ≥8, the number of such subsets is 32/2, or 16.
