1/sin^2x-1/tan^2x=
1/sin^2x-1/ (sin^2x/cos^2x)<<sin tan= sin/cos>>
= 1/sin^2x- cos^2x / sin^2x
= (1- cos^2x) / sin^2x <<combining into a single fraction>>
sin^2 x / sin^2x <<since 1- cos^2 x sin^2 x
=1
this simplifies to 1.
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.
Since 13 can only be divided evenly by 1 and 13 (itself), it is called a prime number.
Answer:
x=0.6241
Step-by-step explanation:
Square both sides and end up with
75.24+x=75.8641
isolate x
x=75.8641 - 75.24
x=0.6241
