Answer:
Let's choose the three odd consecutive numbers 1, 3, and 5.
3^2=9
1•5=5
9-5=4
Let's try the same thing, but with the numbers 101, 103, and 105.
103^2=10609
101•105=10605
10609-10605=4
So, yes, the square of the second number out of three consecutive odd numbers is four greater than the product of the first and the third numbers.
Answer:
y=2/3x-16/3
Step-by-step explanation:
Find perpendular by using negative reciprocal. Then plug in your x and y points into that equation, y=-2/3x-3 --> Then make y=2 and x=-1 --> 2=2/3(-1)+b
solve for b and ur done.
Alright.
For 7, you'll want to put congruent sides equal to each other, assuming they are parallelograms. So, you'll get the two equations:
3x+2=23
2y-7=9
Solve using GEMDAS/PEMDAS, and you'll get these answers.
3x+2=23
3x=21
x=7
2y-7=9
2y=2
y=1
For 8, you'll want to do the exact same thing, formatting the numbers to equal each other. You'll get these two equations:
3y+5=14
2x-5=17
Solving them would make:
3y+5=14
3y=9
y=3
2x-5=17
2x=22
x=11
For 9, you have to remember that the angle opposite of one angle in a defined parallelogram are congruent. Thus:
130=2h
5k=50
solve them and you get
h=65
k=10
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Hope that helped. Good luck.
Answer:
Step-by-step explanation:
Hello. Would love to try and help, your screenshot is not showing up when I click on it though.
