I'm assuming they are equivalent...
![5m + 1 = 3m + 7](https://tex.z-dn.net/?f=5m%20%2B%201%20%3D%203m%20%2B%207)
Solve for m and get 3.
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I hope that helps you out!!
Any more questions, please feel free to ask me and I will gladly help you out!!
~Zoey
Answer: B) y=1/4x+7/2
Step 1: Slope
Knowing that the equation will be perpendicular to the one given will only give us the slope. Perpendicular means that the slope will be opposite to what it is now.
The opposite of -4 (the current slope) is 1/4. Even knowing only this will tell us that the answer is b, but I will continue to explain for when no options are given.
Step 2: Y-intercept
Now that we know the slope, the perpendicular aspect of this equation will no longer be of any help.
To find the y-intercept, we need to substitute y and x with the point given.
Our current equation is— y= 1/4x+b
Let’s substitute the point into this. The point (-2,3) tells us that the y-value is 3 and the x-value is -2.
y= 1/4x+ b
3=1/4(-2)+b
3 = -1/2 +b
+1/2 +1/2
__________
3 1/2 =b
Step 3: Improper fraction
Being that it is preferred to leave fractions improper, let’s turn 3 1/2 into an improper fraction. This will turn into 7/2 (comment below if you need an explanation of how to do this)
Final answer: now that we know that the slope is 1/4 and the y-intercept is 7/2, we know that our equation will be y=1/4x+7/2
Hope this helps comment below for more questions:)
Answer:
45x^2
Step-by-step explanation:
(45+x)(1+x)
how many 3 element subsets of {1, 2, 3, 4, 5, 6, 7, 8, ,9, 10, 11} are there for which the sum of the elements in the subset is
AURORKA [14]
Answer:
There are 155 ways in which these elements casn occur.
Step-by-step explanation:
We want 3 element subsets whose sum are multiples of 3
1+2+3= 6
1+2+6= 9
1+2+9= 12
1+9+11=21
1+3+5=9
1+4+8=12
1+5+6=12
1+6+8=15
1+7+10=18
1+8+9=18
1+9+11=21
2+3+7=12
2+4+6=12
2+4+9=15
2+5+11=18
2+6+7=15
2+7+9=18
2+8+5=15
2+8+11=21
2+9+10=21
3+6+9= 18
3+9+11=21
3+10+11=24
6+9+10=27
6+8+11=27
6+7+11=24
7+8+9= 24
8+9+10=27
7+9+11=27 .........
We have 11 elements
We need a combination of 3
The combinations can be in the form
even+ even+ odd
odd+odd+odd
even + odd+odd
So there are 3 ways in which these elements can occur
Total number of combinations with 3 elements =11C3= 165
There are 6 odd numbers and 5 even numbers.
Number of subsets with 3 odd numbers = 6C3= 20
Number of two even numbers and 1 odd number = 5C2*6C1=10*6= 60
Number of 2 odd and 1 even number = 6C2* 5C1= 5*15= 75
So 20+60+75=155
There are 155 ways in which this combination can occur