First, mentally add the whole numbers.
4+3+5=12
Now, look at the fractions. 1/8, 2/3, and 1/2.
2/3 is bigger than 1/2, so you can add in another whole number.
12+1=13
1/8 is tiny, and won't add a whole lot to the mix, so just leave that.
A reasonable estimate would be 13 1/8
Answers:1)Tthe first answer is that as x increases the value of p(x) approaches a number that is greater than q (x).
2) the y-intercept of the function p is greater than the y-intercept of the function q.
Explanation:1) Value of the functions as x increases.Function p:

As x increases, the value of the function is the limit when x → ∞.
Since [2/5] is less than 1,
the limit of [2/5]ˣ when x → ∞ is 0, and the limit of p(x) is 0 - 3 = -3.While in the graph you see that the function
q has a horizontal asymptote that shows that the
limit of q (x) when x → ∞ is - 4.Then, the first answer is that
as x increases the value of p(x) approaches a number that is greater than q (x).2) y - intercepts.i) To determine the y-intercept of the function p(x), just replace x = 0 in the equation:
p(x) = [ 2 / 5]⁰ - 3 = 1 - 3 = - 2ii) The y-intercept of q(x) is read in the
graph. It is - 3.
Then the answer is that
the y-intercept of the function p is greater than the y-intercept of the function q.
Answer:
(x + 1)² = 7
Step-by-step explanation:
Given:
-2x = x² - 6
We'll start by rearranging it to solve for zero:
x² + 2x - 6 = 0
The first term is already a perfect square so that's fine. Normally, if that term had a non-square coefficient, you would need to multiply all terms a value that would change that constant to a perfect square.
Because it's already square (1), we can simply move to the next step, separating the -6 into a value that can be doubled to give us the 2, the coefficient of the second term. That value will of course be 1, giving us:
x² + 2x + 1 - 1 - 6 = 0
Now can group our perfect square on the left and our constants on the right:
x² + 2x + 1 - 7= 0
x² + 2x + 1 = 7
(x + 1)² = 7
To check our answer, we can solve for x:
x + 1 = ± √7
x = -1 ± √7
x ≈ 1.65, -3.65
Let's try one of those in the original equation:
-2x = x² - 6
-2(1.65) = 1.65² - 6
- 3.3 = 2.72 - 6
-3.3 = -3.28
Good. Given our rounding that difference of 2/100 is acceptable, so the answer is correct.
2,400
20% of 3,000= 600
3,000 - 600= 2,400
(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>