17. The letter is J
The letter appears 3 times
18. Lefthanded students = 4 students
<h3>How to determine the factors</h3>
17. We have that 1/4 of the months of the year start with the same letter
It is important to note that:
January, June and July all start with the same letter
The letter is J
The letter appears 3 times
18. Total number of students = 24
5/ 6 are right handed
Right handed students = 24 × 5/ 6 = 4 × 5 = 20 students
Lefthanded students = 24 - righthanded students
Lefthanded students = 24 - 20
Lefthanded students = 4 students
Thus, we can see that the letter that appear in 3 months of the year is J
Learn more about word problems here:
brainly.com/question/13818690
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first off, make sure you have a Unit Circle, if you don't do get one, you'll need it, you can find many online.
let's double up 67.5°, that way we can use the half-angle identity for the cosine of it, so hmmm twice 67.5 is simply 135°, keeping in mind that 135° is really 90° + 45°, and that whilst 135° is on the 2nd Quadrant and its cosine is negative 67.5° is on the 1st Quadrant where cosine is positive, so
![cos(\alpha + \beta)= cos(\alpha)cos(\beta)- sin(\alpha)sin(\beta) \\\\\\ cos\left(\cfrac{\theta}{2}\right)=\pm \sqrt{\cfrac{1+cos(\theta)}{2}} \\\\[-0.35em] ~\dotfill\\\\ cos(135^o)\implies cos(90^o+45^o)\implies cos(90^o)cos(45^o)~~ - ~~sin(90^o)sin(45^o) \\\\\\ \left( 0 \right)\left( \cfrac{\sqrt{2}}{2} \right)~~ - ~~\left( 1\right)\left( \cfrac{\sqrt{2}}{2} \right)\implies -\cfrac{\sqrt{2}}{2} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=cos%28%5Calpha%20%2B%20%5Cbeta%29%3D%20cos%28%5Calpha%29cos%28%5Cbeta%29-%20sin%28%5Calpha%29sin%28%5Cbeta%29%20%5C%5C%5C%5C%5C%5C%20cos%5Cleft%28%5Ccfrac%7B%5Ctheta%7D%7B2%7D%5Cright%29%3D%5Cpm%20%5Csqrt%7B%5Ccfrac%7B1%2Bcos%28%5Ctheta%29%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20cos%28135%5Eo%29%5Cimplies%20cos%2890%5Eo%2B45%5Eo%29%5Cimplies%20cos%2890%5Eo%29cos%2845%5Eo%29~~%20-%20~~sin%2890%5Eo%29sin%2845%5Eo%29%20%5C%5C%5C%5C%5C%5C%20%5Cleft%28%200%20%5Cright%29%5Cleft%28%20%5Ccfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%5Cright%29~~%20-%20~~%5Cleft%28%201%5Cright%29%5Cleft%28%20%5Ccfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%5Cright%29%5Cimplies%20-%5Ccfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
Step-by-step explanation:
1.) Firstly, you need to count out how many things there are. 4 app, 5 entrees, 2 desserts. This makes 11 different things
2.) for 11 numbers there can be 9 ways for each number for 1st nine places since 0 can't be the fist figure and next two are reciprocated from 10 numbers. Therefore 10!/2!
3.)So 99 combinations have 10!/2! Permutations hence total ways are 99*10!/2!
6 and 1 tenths is 6.1 and six and one tenth.
Answer:
The first time, the ant went 2m away, then we have a₁ = 2m
The second time, the ant went 3m away, then we have a₂ = 3m
The next time, the ant went a distance equal to the sum of the two previous ones, so we can write it in a general case as:
aₙ = aₙ₋₁ + aₙ₋₂
Only if n > 2.
then the third time, we have n = 3.
a₃ = a₂ + a₁ = 2m + 3m = 5m
a₄ = a₃ + a₂ = 3m + 5m = 8m
a₅ = a₄ + a₃ = 5m + 8m = 13m
This means that in the fifth trip, the ant went 13 meters away from its nest.
And if we consider the fact that the ant must go and return, then it will move a total distance of 13 meters two times, or 2*13m = 26 meters.
