Presume y=kx +b,then substitute (3,0)&(5,4),
When x=3,y=0,so 3k+b=0;
When x=5,y=4,so 5k+b=4;
Now u can know that k=2,b=-6.
The answer is:y=2x-6.
The equation in slope-intercept form for the line that passes through the point ( -1 , -2 ) and is perpendicular to the line − 4 x − 3 y = − 5 is ![y = \frac{3}{4}x - \frac{5}{4}](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B3%7D%7B4%7Dx%20-%20%5Cfrac%7B5%7D%7B4%7D)
<em><u>Solution:</u></em>
<em><u>The slope intercept form is given as:</u></em>
y = mx + c ----- eqn 1
Where "m" is the slope of line and "c" is the y - intercept
Given that the line that passes through the point ( -1 , -2 ) and is perpendicular to the line − 4 x − 3 y = − 5
Given line is perpendicular to − 4 x − 3 y = − 5
− 4 x − 3 y = − 5
-3y = 4x - 5
3y = -4x + 5
![y = \frac{-4x}{3} + \frac{5}{3}](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B-4x%7D%7B3%7D%20%2B%20%5Cfrac%7B5%7D%7B3%7D)
On comparing the above equation with eqn 1, we get,
![m = \frac{-4}{3}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B-4%7D%7B3%7D)
We know that product of slope of a line and slope of line perpendicular to it is -1
![\frac{-4}{3} \times \text{ slope of line perpendicular to it}= -1\\\\\text{ slope of line perpendicular to it} = \frac{3}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B-4%7D%7B3%7D%20%5Ctimes%20%5Ctext%7B%20slope%20of%20line%20perpendicular%20to%20it%7D%3D%20-1%5C%5C%5C%5C%5Ctext%7B%20slope%20of%20line%20perpendicular%20to%20it%7D%20%3D%20%5Cfrac%7B3%7D%7B4%7D)
Given point is (-1, -2)
Now we have to find the equation of line passing through (-1, -2) with slope ![m = \frac{3}{4}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B3%7D%7B4%7D)
Substitute (x, y) = (-1, -2) and m = 3/4 in eqn 1
![-2 = \frac{3}{4}(-1) + c\\\\-2 = \frac{-3}{4} + c\\\\c = - 2 + \frac{3}{4}\\\\c = \frac{-5}{4}](https://tex.z-dn.net/?f=-2%20%3D%20%5Cfrac%7B3%7D%7B4%7D%28-1%29%20%2B%20c%5C%5C%5C%5C-2%20%3D%20%5Cfrac%7B-3%7D%7B4%7D%20%2B%20c%5C%5C%5C%5Cc%20%3D%20-%202%20%2B%20%5Cfrac%7B3%7D%7B4%7D%5C%5C%5C%5Cc%20%3D%20%5Cfrac%7B-5%7D%7B4%7D)
![\text{ substitute } c = \frac{-5}{4} \text{ and } m = \frac{3}{4} \text{ in eqn 1}](https://tex.z-dn.net/?f=%5Ctext%7B%20substitute%20%7D%20c%20%3D%20%5Cfrac%7B-5%7D%7B4%7D%20%5Ctext%7B%20and%20%7D%20m%20%3D%20%5Cfrac%7B3%7D%7B4%7D%20%5Ctext%7B%20in%20eqn%201%7D)
![y = \frac{3}{4} \times x + \frac{-5}{4}\\\\y = \frac{3}{4}x - \frac{5}{4}](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7B3%7D%7B4%7D%20%5Ctimes%20x%20%2B%20%5Cfrac%7B-5%7D%7B4%7D%5C%5C%5C%5Cy%20%3D%20%5Cfrac%7B3%7D%7B4%7Dx%20-%20%5Cfrac%7B5%7D%7B4%7D)
Thus the required equation of line is found
Here you are giving only the amount they want to raise (namely profit times number of magazines sold), and here you are also giving Money they want to raise... So clarifying, the money they want to raise, should include the money they will spend on buying the magazines (there is no statement saying they found them, or were given the magazines, so a cost should be involved)
Now if they are only making the count of "Field trip costs X amount of money, and given we have to make a profit of $5.5, How many must we sell?" then the equation should be n=X/5.5
Should the story be, how much money must they raise to have a profit of 5.5 on each magazine and still have enough for the field trip, then you have a different equation which varies only in adding the cost of each magazine, either case, M should be defined not as money they need to raise (cause here they will be short on their goal) but Money they must earn. And again, you should rewrite your equation to be:
M=Amount they must raise
C=Cost per magazine
n=Number of magazines
p=profit $5.5 per magazine
C+p=M/n
And rewriting the previous they should make:
n(C+p)=M -----> n(C+5.5)=M <span>m/n = 5.50 </span>
<span>m/n x n = 5.50 x n //// multiply each side by n </span>
<span>m = 5.5n</span>
Answer:
12
Step-by-step explanation:
in this problem there is A+b=15
A=9
B=12
Answer:
no
Step-by-step explanation: