(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>
Answer:
The standard form is 8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11
The degree of given polynomial is '5'
the co-efficient of y⁴ is '-17'
Step-by-step explanation:
Given standard form 2 y²+ 6 y³-11-17 y⁴+8 y⁵
<em>The form ax² + b x + c is called the standard form of the quadratic expression of 'x'.This is second degree standard form of polynomial.</em>
<em>The form ax⁵ + b x⁴ + c x³ +d x² +ex +f is called the standard form of the quadratic expression of 'x'.This is fifth degree standard form of polynomial</em>
now Given polynomial is 2 y²+ 6 y³-11-17 y⁴+8 y⁵
The standard form is
8 y ⁵ - 17 y⁴ + 6 y³ +2 y² - 11
<u><em>Conclusion</em></u>:-
<em>The degree of given polynomial is '5'</em>
<em>The co-efficient of y⁴ is '-17'</em>
<em> </em>
Hi there!
Since the total weight the box can hold is 8 pounds, and she filled up 2/4:
8 * 2/4 = 4
There were 4 pounds in the box.
Hope this helps!
Answer is 
dm².
In order to calculate the area of the board that can be seen, we need to calculate the area of the the rectangular board and the circular frame. The difference of both's area is the required answer.
Refer to the attachment for step by step solution. :)
The correct question is: evaluate
![\sqrt[3]{- \frac{8}{125} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B-%20%5Cfrac%7B8%7D%7B125%7D%20%7D%20)
The
answer is: option B. -2 / 5Explanation:![\sqrt[3]{- \frac{8}{125} }=- \frac{ \sqrt[3]{8} }{ \sqrt[3]{125} } =- \frac{ \sqrt[3]{2^3} }{ \sqrt[3]{5^3} } =- \frac{2}{5}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B-%20%5Cfrac%7B8%7D%7B125%7D%20%7D%3D-%20%5Cfrac%7B%20%5Csqrt%5B3%5D%7B8%7D%20%7D%7B%20%5Csqrt%5B3%5D%7B125%7D%20%7D%20%3D-%20%5Cfrac%7B%20%5Csqrt%5B3%5D%7B2%5E3%7D%20%7D%7B%20%5Csqrt%5B3%5D%7B5%5E3%7D%20%7D%20%3D-%20%5Cfrac%7B2%7D%7B5%7D%20)
