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
___________________________________________
Hope that helped. Good luck.
Answer:
17.15
Step-by-step explanation:
for example lets use 3^2 + 4^2, you add the main numbers, 3 and 4 to get 7, then there are still the exponents, the squared signs, they don't change, they stay as a two. If your problem required a single whole number, then you have to square the 7, 7*7 is 49, so that or 7^2 would be your final answer.
QUESTION 1
The given system of equations are:
We equate the two equations to get:
When x=0,
The solution is (0,1)
QUESTION 2
The given equations are:
and
We equate both equations to get:
Group similar terms,
We put x=0 into any of the equations to find y.
The solution is (0,-1).
QUESTION 3
The given equations are:
and
We equate both equations:
Group similar terms:
This is not true.
Hence the system has no solution.
Answer:
Option (2)
Step-by-step explanation:
In this question we have to find the multiplication of the two expressions.
(2p + q)(-3q - 6p + 1)
= 2p(-3q - 6p + 1) + q(-3q - 6p + 1) [By distributive property]
= -6pq - 12p²+ 2p - 3q² - 6pq + q
= -12p² - (6pq + 6pq) - 3q² + 2p + q
= -12p² - 12pq + 2p - 3q² + q
Therefore, Option (2) will be the correct option.
