In both problems, the sum of side lengths is the perimeter. Opposite sides of a parallelogram (or rectangle) are equal in length, so you can find the perimeter by doubling the sum of adjacent sides.
25. 2(x +(x +15)) = (x +45) +(x +40) +(x +25)
.. 4x +30 = 3x +110 . . . . . . . . . . . . . . . . . . . . . . simplify
.. x = 80 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . subtract 3x+30
.. 4x +30 = 4*80 +30 = 240
The perimeter of each is 240 units.
26. 2(x +(x +2)) = (x) +(x +6) +(x +4)
.. 4x +4 = 3x +10 . . . . . . . . . . . . . . . . . . . . . . simplify
.. x = 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . subtract 3x+4
.. 4x +4 = 4*6 +4 = 28
The perimeter of each is 28 units.
Answer:
becauz the cos .1sj sin A the thing cos
Answer:
The pairs are (13,15) and (-15,-13).
Step-by-step explanation:
If n is an odd integer, the very next odd integer will be n+2.
n+1 is even (so we aren't using this number)
The sum of the squares of (n) and (n+2) is 394.
This means
(n)^2+(n+2)^2=394
n^2+(n+2)(n+2)=394
n^2+n^2+4n+4=394 since (a+b)(a+b)=a^2+2ab+b^2
Combine like terms:
2n^2+4n+4=394
Subtract 394 on both sides:
2n^2+4n-390=0
Divide both sides by 2:
n^2+2n-195=0
Now we need to find two numbers that multiply to be -195 and add up to be 2.
15 and -13 since 15(-13)=-195 and 15+(-13)=2
So the factored form is
(n+15)(n-13)=0
This means we have n+15=0 and n-13=0 to solve.
n+15=0
Subtract 15 on both sides:
n=-15
n-13=0
Add 13 on both sides:
n=13
So if n=13 , then n+2=15.
If n=-15, then n+2=-13.
Let's check both results
(n,n+2)=(13,15)
13^2+15^2=169+225=394. So (13,15) looks good!
(n,n+2)=(-15,-13)
(-15)^2+(-13)^2=225+169=394. So (-15,-13) looks good!
2 and 3 are both parallel because both slopes are m=3/5, which equal slopes-parallel lines
Answer:
x=16°
Step-by-step explanation:
tanθ=opposite/adjacent
tan(<CAB)=BC/AB
<CAB=tan⁻¹(21/29)=35.9097308°
tanθ=opposite/adjacent
tan(<DAB)=BD/AB
<DAB=tan⁻¹((21/2)/29)=19.90374954°
<CAB=<DAB+x
x=<CAB-<DAB
=35.9097308-19.90374954
x=16.00598126°
