Answer:
z = 17.5
Step-by-step explanation:
segment RX is an angle bisector and divides the opposite side into segments that are proportional to the other 2 sides, that is
= 
, substitute values
= 
( cross- multiply )
6z = 105 ( divide both sides by 6 )
z = 17.5
Given the equation - x² + 5x = 3, which can be rewritten as:
- x² + 5x - 3 = 0
where a = -1, b = 5 and c = -3.
Quadratic formula:
![\frac{-b\text{ }\pm\text{ }\sqrt[]{b^2\text{ - 4ac}}}{2a}](https://tex.z-dn.net/?f=%5Cfrac%7B-b%5Ctext%7B%20%7D%5Cpm%5Ctext%7B%20%7D%5Csqrt%5B%5D%7Bb%5E2%5Ctext%7B%20-%204ac%7D%7D%7D%7B2a%7D)
Now, we just replace the values of a, b and c on the equation above.
![\frac{-5\text{ }\pm\text{ }\sqrt[]{5^2\text{ - 4(-1)(3)}}}{2(-1)}](https://tex.z-dn.net/?f=%5Cfrac%7B-5%5Ctext%7B%20%7D%5Cpm%5Ctext%7B%20%7D%5Csqrt%5B%5D%7B5%5E2%5Ctext%7B%20-%204%28-1%29%283%29%7D%7D%7D%7B2%28-1%29%7D)
=
![\frac{5}{2}\text{ }\pm\text{ }\frac{\sqrt[]{13}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B5%7D%7B2%7D%5Ctext%7B%20%7D%5Cpm%5Ctext%7B%20%7D%5Cfrac%7B%5Csqrt%5B%5D%7B13%7D%7D%7B2%7D)
(e) Each license has the formABcxyz;whereC6=A; Bandx; y; zare pair-wise distinct. There are 26-2=24 possibilities forcand 10;9 and 8 possibilitiesfor each digitx; yandz;respectively, so that there are 241098 dierentlicense plates satisfying the condition of the question.3:A combination lock requires three selections of numbers, each from 1 through39:Suppose that lock is constructed in such a way that no number can be usedtwice in a row, but the same number may occur both rst and third. How manydierent combinations are possible?Solution.We can choose a combination of the formabcwherea; b; carepair-wise distinct and we get 393837 = 54834 combinations or we can choosea combination of typeabawherea6=b:There are 3938 = 1482 combinations.As two types give two disjoint sets of combinations, by addition principle, thenumber of combinations is 54834 + 1482 = 56316:4:(a) How many integers from 1 to 100;000 contain the digit 6 exactly once?(b) How many integers from 1 to 100;000 contain the digit 6 at least once?(a) How many integers from 1 to 100;000 contain two or more occurrencesof the digit 6?Solutions.(a) We identify the integers from 1 through to 100;000 by astring of length 5:(100,000 is the only string of length 6 but it does not contain6:) Also not that the rst digit could be zero but all of the digit cannot be zeroat the same time. As 6 appear exactly once, one of the following cases hold:a= 6 andb; c; d; e6= 6 and so there are 194possibilities.b= 6 anda; c; d; e6= 6;there are 194possibilities. And so on.There are 5 such possibilities and hence there are 594= 32805 such integers.(b) LetU=f1;2;;100;000g:LetAUbe the integers that DO NOTcontain 6:Every number inShas the formabcdeor 100000;where each digitcan take any value in the setf0;1;2;3;4;5;7;8;9gbut all of the digits cannot bezero since 00000 is not allowed. SojAj= 9<span>5</span>
