In the first column you put 40 50 and 60
Then in the 2nd column you put those bombers into the equation (so you would write 3(40)+2 3(50)+2 and 3(60)+2) and then in the third column you solve them (so 3(40)+2=122 3(50)+2=152 3(60)+2=182)
And then you graph them at (40,122) (50,152) and (60,182)
Step-by-step explanation:
x > -1
x = {0,1,2,...}
![\lang -1,+\infty \rang](https://tex.z-dn.net/?f=%20%5Clang%20-1%2C%2B%5Cinfty%20%5Crang)
Answer:
$5.35
Step-by-step explanation:
.85+4.50= 5.35
25c(quarters) x 12 = $3.00
If theres 28 coins, and 12 are quarters, then that means theres 16 coins left to use for dimes.
10c(dimes) x 16 = $1.60
$3.00 + $1.60 = $4.60
Therefore, there are 12 quarters and 16 dimes.
Hope this helps! Feel free to ask any questions.
In order to check if the lines are perpendicular, we need to check if their slopes have the following relation (to find the slope we can use the slope-intercept form y = mx + b):
![m_1=-\frac{1}{m_2}](https://tex.z-dn.net/?f=m_1%3D-%5Cfrac%7B1%7D%7Bm_2%7D)
A.
In this option, y = -5 is an horizontal line and x = 2 is a vertical line, therefore they are perpendicular.
B.
First let's find the slope of each line:
![\begin{gathered} x+\frac{y}{5}=2 \\ 5x+y=10 \\ y=-5x+10\to m=-5 \\ \\ -\frac{x}{5}+y=3 \\ y=\frac{x}{5}+3\to m=\frac{1}{5} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20x%2B%5Cfrac%7By%7D%7B5%7D%3D2%20%5C%5C%205x%2By%3D10%20%5C%5C%20y%3D-5x%2B10%5Cto%20m%3D-5%20%5C%5C%20%20%5C%5C%20-%5Cfrac%7Bx%7D%7B5%7D%2By%3D3%20%5C%5C%20y%3D%5Cfrac%7Bx%7D%7B5%7D%2B3%5Cto%20m%3D%5Cfrac%7B1%7D%7B5%7D%20%5Cend%7Bgathered%7D)
These slopes obey the relation stated above, so the lines are perpendicular.
C.
![\begin{gathered} y=\frac{1}{3}x+1\to m=\frac{1}{3} \\ \\ y-1=-3(x-5) \\ y-1=-3x+15 \\ y=-3x+16\to m=-3 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%3D%5Cfrac%7B1%7D%7B3%7Dx%2B1%5Cto%20m%3D%5Cfrac%7B1%7D%7B3%7D%20%5C%5C%20%20%5C%5C%20y-1%3D-3%28x-5%29%20%5C%5C%20y-1%3D-3x%2B15%20%5C%5C%20y%3D-3x%2B16%5Cto%20m%3D-3%20%5Cend%7Bgathered%7D)
These slopes obey the relation stated above, so the lines are perpendicular.
D.
![\begin{gathered} y-5=x+1 \\ y=x+6\to m=1 \\ \\ x+y=3 \\ y=-x+3\to m=-1 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y-5%3Dx%2B1%20%5C%5C%20y%3Dx%2B6%5Cto%20m%3D1%20%5C%5C%20%20%5C%5C%20x%2By%3D3%20%5C%5C%20y%3D-x%2B3%5Cto%20m%3D-1%20%5Cend%7Bgathered%7D)
These slopes obey the relation stated above, so the lines are perpendicular.
E.
![\begin{gathered} y+2=\frac{1}{3}(x-6) \\ y+2=\frac{1}{3}x-2 \\ y=\frac{1}{3}x-4\to m=\frac{1}{3} \\ \\ y=3x+4\to m=3 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20y%2B2%3D%5Cfrac%7B1%7D%7B3%7D%28x-6%29%20%5C%5C%20y%2B2%3D%5Cfrac%7B1%7D%7B3%7Dx-2%20%5C%5C%20y%3D%5Cfrac%7B1%7D%7B3%7Dx-4%5Cto%20m%3D%5Cfrac%7B1%7D%7B3%7D%20%5C%5C%20%20%5C%5C%20y%3D3x%2B4%5Cto%20m%3D3%20%5Cend%7Bgathered%7D)
These slopes don't obey the relation stated above, so the lines aren't perpendicular.
The correct options are A, B, C and D.