The one you have highlighted is correct
You have to add 75 to Both sides and 50+75= 125
So q=125. Hope this helped!
In order to write this quadratic equation in standard form, first note that standard form is ax^2+bx+c for quadratics, where c is the numerical value (constant), B is the coefficient of x, and a is the coefficient of x^2 and is the leading coefficient. Next, multiply the binomials of (x-7) and (x-1). You can do this by using FOIL, or by distributing each of the terms in a binomial to each of the other terms in the other binomial. (Please let me know if you need a walk through in this step in particular). Furthermore, you should then write y= (the simplified trinomial). Now, the quadratic is in standard form. To reiterate, just simplify the two binomials by multiplying them together and writing that they're equal to y.
M(x)=−(x−5) 2 +25m, left parenthesis, x, right parenthesis, equals, minus, left parenthesis, x, minus, 5, right parenthesis, squ
Aleksandr [31]
Answer:
25
Step-by-step explanation:
Given the function that represents the amount of mosquitoes as;
M(x) = -(x-5)² + 25
To get the maximum mosquitoes possible, we need to first find x;
At the maximum dM/dx = 0
dM//dx = -2(x-5)
0 = -2(x-5)
-2(x-5) = 0
-2x + 10 = 0
-2x = -10
x = 10/2
x = 5
Substitute x = 5 into the function;
M(5) = −(5−5)² +25
M(5) = 0+25
M(5) = 25
Hence the maximum number of mosquitoes is 25
Check the picture below.
well, we want only the equation of the diametrical line, now, the diameter can touch the chord at any several angles, as well at a right-angle.
bearing in mind that <u>perpendicular lines have negative reciprocal</u> slopes, hmm let's find firstly the slope of AB, and the negative reciprocal of that will be the slope of the diameter, that is passing through the midpoint of AB.
![\bf A(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad B(\stackrel{x_2}{5}~,~\stackrel{y_2}{1}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{1}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{5}-\underset{x_1}{1}}}\implies \cfrac{-3}{4} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{slope of AB}}{-\cfrac{3}{4}}\qquad \qquad \qquad \stackrel{\textit{\underline{negative reciprocal} and slope of the diameter}}{\cfrac{4}{3}}](https://tex.z-dn.net/?f=%5Cbf%20A%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20B%28%5Cstackrel%7Bx_2%7D%7B5%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%20~%5Chfill%20%5Cstackrel%7Bslope%7D%7Bm%7D%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%20%7B%5Cstackrel%7By_2%7D%7B1%7D-%5Cstackrel%7By1%7D%7B4%7D%7D%7D%7B%5Cunderset%7Brun%7D%20%7B%5Cunderset%7Bx_2%7D%7B5%7D-%5Cunderset%7Bx_1%7D%7B1%7D%7D%7D%5Cimplies%20%5Ccfrac%7B-3%7D%7B4%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bslope%20of%20AB%7D%7D%7B-%5Ccfrac%7B3%7D%7B4%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7B%5Cunderline%7Bnegative%20reciprocal%7D%20and%20slope%20of%20the%20diameter%7D%7D%7B%5Ccfrac%7B4%7D%7B3%7D%7D)
so, it passes through the midpoint of AB,
so, we're really looking for the equation of a line whose slope is 4/3 and runs through (3 , 5/2)
