Answer:
I think its choice C
Step-by-step explanation:
Because it makes more sense to me
5 * 1.5 to get 7.5, which is total pounds
7.5 - 2.8 to get how many almonds
= 4.7
Choice C makes the most sense to get that answer
i'm only in 5th grade so tell me if i'm wrong ill be ok with that
Answer:
Total amount of fencing needed as an algebraic expression in terms of x is: <em>10x</em><em> </em><em>+</em><em> </em><em>3</em> .
Step-by-step explanation:
As it is given that each rectangle has the same dimensions, the dimensions of each rectangle must be: x units by 2x + 1 units.
Based on this, we can calculate the total amount of fencing needed.
Let width of each rectangle = x
Let length of each rectangle = 2x + 1
There are 4 widths and 3 lengths in total of fencing.
Therefore:
= 4 ( x ) + 3 ( 2x + 1 )
Expand:
= 4x + 6x + 3
Group like-terms:
= 10x + 3
Answer:
144 ways
Step-by-step explanation:
Number of paintings = 7
Renaissance = 4
Baroque = 3
We are hanging from left to right and we will first hang Renaissance painting before baroque painting.
For Renaissance we have 4! Ways of doing so. 4 x3x2x1 = 24
For baroque we have 3! Ways of doing so. 3x2x1 = 6
We have 4!ways x 3!ways
= (4x3x2x1) * (3x2x1) ways
= 144 ways
Therefore we have 144 ways to hang the painting.
Answer:
y(t) = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t) + 1
Step-by-step explanation:
y" + y' + y = 1
This is a second order nonhomogenous differential equation with constant coefficients.
First, find the roots of the complementary solution.
y" + y' + y = 0
r² + r + 1 = 0
r = [ -1 ± √(1² − 4(1)(1)) ] / 2(1)
r = [ -1 ± √(1 − 4) ] / 2
r = -1/2 ± i√3/2
These roots are complex, so the complementary solution is:
y = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t)
Next, assume the particular solution has the form of the right hand side of the differential equation. In this case, a constant.
y = c
Plug this into the differential equation and use undetermined coefficients to solve:
y" + y' + y = 1
0 + 0 + c = 1
c = 1
So the total solution is:
y(t) = c₁ e^(-1/2 t) cos(√3/2 t) + c₂ e^(-1/2 t) sin(√3/2 t) + 1
To solve for c₁ and c₂, you need to be given initial conditions.
