3421880 number of the different meals are possible.
<h3>What is the combination?</h3>
The arrangement of the different things or numbers in the number of the ways is called as the combination.
It is given that:-
Number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts.
So let's take each course by itself. You can choose 1 of 7 appetizers. So we have n = 7 After that, you chose an entre,
so the number of possible meals to this point is
n = 7 x 10 = 70
Finally, you finish off with a dessert, so the number of meals is:
n = 70 x 4 = 280
Therefore the number of possible meals you can have is 280. Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is
n = 77 x 1010 x 44 = 3421880
Therefore 3421880 different meals are possible.
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Look how it is done and you will understand
Answer:
∠1=∠9=43°, ∠2=∠7= 17°, ∠4=∠5=∠6=60°, ∠3=∠8=120°
Step-by-step explanation:
Given ABC is an isosceles triangle and DBE is an equilateral triangle. we have to find each missing measure.
In triangle ABC,
∠1=∠9 (∵isosceles triangle)
⇒ 4x+3=9x-47
⇒ 9x-4x=3+47 ⇒ 5x=50 ⇒ x=10
Hence, ∠9=∠1=(4x+3)°=4(10)+3=43°
Also, given ΔBDE is an equilateral triangle, and all angle of equilateral triangle are equal.
∴ ∠4+∠5+∠6=180°
⇒ ∠4+∠4+∠4=180° ⇒ 3∠4 = 180° ⇒ ∠4 = 60°
∴ ∠4=∠5=∠6=60°
By exterior angle property, ∠3=∠5+∠6=60°+60°=120°
∠8=∠5+∠4=60°+60°=120°
In ΔABD, ∠1+∠2+∠3=180°
⇒ 43°+120°+∠2=180°⇒ ∠2 = 17°
In ΔABD, ∠9+∠8+∠7=180°
⇒ 43°+120°+∠7=180°⇒ ∠7= 17°
Step-by-step explanation:
Answer:
4.913
Step-by-step explanation:
T=2π/|b|. The period of an equation of the form y = a sin bx is T=2π/|b|.
In mathematics the curve that graphically represents the sine function and also that function itself is called sinusoid or sinusoid. It is a curve that describes a repetitive and smooth oscillation. It can be represented as y(x) = a sin (ωx+φ) where a is the amplitude, ω is the angular velocity with ω=2πf, (ωx+φ) is the oscillation phase, and φ the initial phase.
The period T of the sin function is T=1/f, from the equation ω=2πf we can clear f and substitute in T=1/f.
f=ω/2π
Substituting in T=1/f:
T=1/ω/2π -------> T = 2π/ω
For the example y = a sin bx, we have that a is the amplitude, b is ω and the initial phase φ = 0. So, we have that the period T of the function a sin bx is:
T=2π/|b|
