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nalin [4]
3 years ago
13

Find the equation of the line Use exact numbers y=*blank*x+*blank*

Mathematics
1 answer:
Murljashka [212]3 years ago
6 0
You are looking for the slope and y intercept to complete the equation of the line.

The equation of a line is in something called slope intercept form. That looks like y = mx + b. m represents the slope (measure of how steep a line is, and in which direction it is going) and b represents the y intercept (y coordinate when x = 0). You need to find the slope and y intercept to complete the equation.

First, find the slope. The formula for slope is: m = (y2 - y1)/(x2 - x1) where m is the slope and (x1, y1) and (x2, y2) are points.

Pick any two points on the graph. I will use (-2, 0) and (0, 4). Now use these values to find the slope.

m = (4-0)/(0+2) = 4/2
m = 2

m = 2 means that for every two units the line goes up on the y axis, it moves one to the right on the x axis. 2 will go in your first box.

Now find the y intercept. The y intercept is where the line crosses the y axis - it is the y coordinate when x = 0. Here when x = 0, y = 4, so your y intercept is at 4. 4 goes into your second box.

The equation is y = 2x + 4

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natta225 [31]
<span>75%=.75
Convert to a  fraction
0.75 = 3/4
(3/4) of 4, that means (3/4) times 4. (3/4)*4 = 12/4 = 3
3 is </span>the answer
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 find all possible value of the given variable 
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1.\\ \\ h^2+5h=0 \\ \\h(x+5)=0\\ \\x=0 \ \ \ or \ \ \ x+5 =0\ \ |-5\\ \\x+5-5=0-5\\ \\x=0 \ \ \ or \ \ \ x=-5


2.\\ \\ z^2-z=0\\ \\z(x-1)=0\\ \\z=0 \ \ \ or \ \ \ z-1 =0 \ \ | +1\\ \\z-1+1 =0 +1 \\ \\x=0 \ \ \ or \ \ \ z=1


3.\\ \\m^2+13m+40=0 \\ \\a=1 ,\ b=13, \ c=40 \\ \\\Delta =b^2-4ac =13^2-4\cdot 1\cdot 40=169 - 1600=-1431 \\ \\and \ we \ know \ when \ \Delta \ is \ negative, \ theres \ no \solution


4.\\ \\z^2-3z=0 \\ \\ (z-3)=0\\ \\z=0 \ \ \ or \ \ \ z-3 =0\ \ |+3\\ \\ z-3+3=0+3\\ \\z=0 \ \ \ or \ \ \ z=3


5.\\ \\q^2+7q=0 \\ \\q(q+7)=0\\ \\q=0 \ \ \ or \ \ \ q+7 =0\ \ |-7\\ \\q+7-7=0-7\\ \\q=0 \ \ \ or \ \ \ q=-7


6.\\ \\k^2+2k=0\\ \\k(k+2)=0\\ \\k=0 \ \ \ or \ \ \ k+2 =0\ \ |-2\\ \\k+2-2=0-2\\ \\k=0 \ \ \ or \ \ \ k=-2


7. \\ \\ x^2-3x-70=0 \\ \\a=1,\ b=-3, \ c=-70 \\ \\\Delta =b^2-4ac = (-3)^2-4\cdot 1\cdot (-70)= 9+280=289\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{3-\sqrt{289}}{2 }=\frac{ 3-17}{2}=\frac{-14}{2}=-7

x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{3+\sqrt{289}}{2 }=\frac{ 3+17}{2}=\frac{20}{2}=10\\ \\(x+7)(x-10)=0


8.\\ \\q^2+7q-60=0 \\ \\a=1,\ b=7, \ q=-60 \\ \\\Delta =b^2-4ac = 7^2-4\cdot 1\cdot (-60)=49+240=289 \\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-7-\sqrt{289}}{2 }=\frac{ -7-17}{2}=\frac{-24}{2}=-12

x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-7+\sqrt{289}}{2 }=\frac{ -7+17}{2}=\frac{ 10}{2}= 5\\ \\(x+12)(x-5)=0


9.\\ \\z^2+9z-36=0 \\ \\a=1,\ b=9, \ q=-36 \\ \\\Delta =b^2-4ac = 9^2-4\cdot 1\cdot (-36)= 81+144=225\\ \\ x_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{-9-\sqrt{225}}{2 }=\frac{ -9-15}{2}=\frac{-24}{2}=-12

x_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{-9+\sqrt{225}}{2 }=\frac{ -9+15}{2}=\frac{6}{2}=3\\ \\(x+11)(x-3)=0


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d_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{13+\sqrt{81}}{2 }=\frac{ 13+9}{2}=\frac{22}{2}=11\\ \\(d-2)(d-11)=0


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STatiana [176]
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emelia earns $8.74 per hour plus a gas allowence of $3.50 per day at her job how much does Emelia jobs pays in a day when she wo
trapecia [35]

Answer:

8.74 (3.50) + 3.50

= $51.57

Step-by-step explanation:

8 0
2 years ago
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