Area of triangle=1/2(base×height)
360=1/2((4x-2)(7x+5))
720=(4x-2)(7x+5)
FOIL the binomial
720=28x^2+6x-10
Bring 720 to the other side
28x^2+6x-10-720=0
28x^2+6x-730=0
Factor 2 out
2(14x^2+3x-365)=0
Factor 2(14x+73)(x-5)=0
x-5=0
x=5
It can't be 14x-73=0 because the length can't be negative.
You're welcome.
Answer:
(3, 50) and (14,303)
Step-by-step explanation:
Given the system of equations;
y=23x–19 ....1
x²–y= – 6x–23 ...2
Substitute 1 into 2;
x²–(23x-19)= – 6x–23
x²–23x+19= – 6x–23 .
x²-23x + 6x + 19 + 23 = 0
x² - 17x + 42 = 0
Factorize;
x² - 14x - 3x + 42 = 0
x(x-14)-3(x-14) = 0
(x-3)(x-14) = 0
x = 3 and 14
If x = 3
y = 23(3) - 19
y = 69-19
y = 50
If x = 14
y = 23(14) - 19
y = 322-19
y = 303
Hence the coordinate solutions are (3, 50) and (14,303)
(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>
Fgh is 144 because 360 - 216 = 144
