Find a, b, and n of the factored form of the binomial expansion.
1 answer:
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A right triangle will be formed with 2.2 km as the hypotenuse, 0.5 degrees as the angle, and an unknown height (see attachment).
Recall the trigonmetric ratios sine, cosine, and tangent that can be used for right triangles.
Sin<span>θ = opposite side/hypotenuse
</span>Cos<span>θ = adjacent side/hypotenuse
</span>tan<span>θ = opposite side / adjacent side
</span>θ (theta) refers to the angle (not the right angle). In this case, θ is the given angle of 0.5°.
In this problem, you will use sine because there is a side opposite to the angle (the unknown side h) and the hypotenuse, and sine relates those two sides.
Let the variable h represent the unknown opposite side, the height.
sin(0.5°) = h / (2.2km)
h = (2.2km)sin(0.5<span>°)
h = 0.019 km
Hope I helped! If you have questions on any of the steps, please comment below.</span>
Recall the trigonmetric ratios sine, cosine, and tangent that can be used for right triangles.
Sin<span>θ = opposite side/hypotenuse
</span>Cos<span>θ = adjacent side/hypotenuse
</span>tan<span>θ = opposite side / adjacent side
</span>θ (theta) refers to the angle (not the right angle). In this case, θ is the given angle of 0.5°.
In this problem, you will use sine because there is a side opposite to the angle (the unknown side h) and the hypotenuse, and sine relates those two sides.
Let the variable h represent the unknown opposite side, the height.
sin(0.5°) = h / (2.2km)
h = (2.2km)sin(0.5<span>°)
h = 0.019 km
Hope I helped! If you have questions on any of the steps, please comment below.</span>
Answer:
0.1 or 1/10
Step-by-step explanation:
The digits 4, 5, 6, 7 and 8 are randomly arranged to form a three digit number, where the digits are not repeated.
This is question of permutation.
Imagine this sum as; there are 3 boxes(blank spaces for digits) and 5 different fruits(digits) are to be put in these boxes, where a box can hold a maximum of only 1 fruit. The number of such permutations are: ⁵P₃
By formula (a! is factorial a):
ᵃPₙ
⁵P₃
⁵P₃
⁵P₃
⁵P₃= 60
This is the total count of possible numbers that can be formed.
Now, for a number to be greater than 800 and even; first digit should necessarily be 8. Last digit can be 4 or 6. Using these conditions, there are 6 possibilities. 854, 864, 874, 846, 856, 876 are the numbers.
The probability that number is even and greater than 800 is: