64005874
word form: sixty-four million, five thousand, eight hundred and seventy-four
expanded form: 60000000 + 4000000 + 5000 + 800 + 70 + 4
30679100
word: thirty million, six hundred seventy nine thousand, and one hundred
30000000 + 600000 + 70000 + 9000 + 100
hope this helps
Answer:
C. YES
Step-by-step explanation:
If two angles and the non-included side of one triangle are equal to the corresponding angles and side of another triangle, the triangles are congruent.
a)
(59, 76) and (65, 100)
y - y1 = m(x - x1)
y - 100 = (100-76)/(65 - 59)(x - 65)
y - 100 = (24/6)(x - 65)
y - 100 = 4(x - 65)
y - 100 = 4x - 260
y = 4x - 160
b)
x = 90
y = 4x - 160
y = 4(90) - 160
y = 360 - 160
y = 200
200 chirps per minute
c)
y = 120
y = 4x - 160
120 = 4x - 160
280 = 4x
x = 70
70 degrees
A volleyball uniform costs $13 for the shirt, $11 for pants, and $8 for socks. Write two equivalent expressions for the total cost of 12 uniforms. Then find the cost.
Write an expression for the cost of 1 uniform, add $13, $11 and $8 .
$13+$11+$8
Write an expression for the cost of 12 uniforms, multiply 12 by the cost of 1 uniform.
12($13+$11+$8)
Simplify using the distributive property.
=12⋅13+12⋅11+12⋅8=156+132+96
Now Add.
=384
So, the total cost of 12 uniforms is $384 .
Answer:
a. L{t} = 1/s² b. L{1} = 1/s
Step-by-step explanation:
Here is the complete question
The The Laplace Transform of a function ft), which is defined for all t2 0, is denoted by Lf(t)) and is defined by the improper integral Lf))s)J" e-st . f(C)dt, as long as it converges. Laplace Transform is very useful in physics and engineering for solving certain linear ordinary differential equations. (Hint: think of s as a fixed constant) 1. Find Lft) (hint: remember integration by parts) A. None of these. B. O C. D. 1 E. F. -s2 2. Find L(1) A. 1 B. None of these. C. 1 D.-s E. 0
Solution
a. L{t}
L{t} = ∫₀⁰⁰
Integrating by parts ∫udv/dt = uv - ∫vdu/dt where u = t and dv/dt =
and v =
and du/dt = dt/dt = 1
So, ∫₀⁰⁰udv/dt = uv - ∫₀⁰⁰vdu/dt w
So, ∫₀⁰⁰
= [
]₀⁰⁰ - ∫₀⁰⁰
∫₀⁰⁰
= [
]₀⁰⁰ - ∫₀⁰⁰
= -1/s(∞exp(-∞s) - 0 × exp(-0s)) +
[
]₀⁰⁰
= -1/s[(∞exp(-∞) - 0 × exp(0)] - 1/s²[exp(-∞s) - exp(-0s)]
= -1/s[(∞ × 0 - 0 × 1] - 1/s²[exp(-∞) - exp(-0)]
= -1/s[(0 - 0] - 1/s²[0 - 1]
= -1/s[(0] - 1/s²[- 1]
= 0 + 1/s²
= 1/s²
L{t} = 1/s²
b. L{1}
L{1} = ∫₀⁰⁰
= [
]₀⁰⁰
= -1/s[exp(-∞s) - exp(-0s)]
= -1/s[exp(-∞) - exp(-0)]
= -1/s[0 - 1]
= -1/s(-1)
= 1/s
L{1} = 1/s