Answer: x = 2
Step-by-step explanation:
First, Distribute
-2+8x=3x+8
Then, Subtract 3x
-2 + 5x=8
Then, Add 2
5x=10
Then, Divide by 5
x=2
Hope it helps <3
x = 44°
Step-by-step explanation:
Since AB is a diameter,
arcAC + arcCB = 180
92° + arcCB = 180
or
arcCB = 88°
ArcCB is the intercepted arc and by definition, the inscribed angle x is half the measure of the intercepted arc. Therefore,
x = (1/2)arcCB
= 44°
Answer:
0.808 kilometers
Step-by-step explanation:
The LCD is basically the LCM of the denoenator so
to find LCM of exg 6 and 21, find factors and group
6=2 and 3 (one 2 and one 3)
21=3 and 7 (one 3 and one 7)
lcm=2 and 3 and 7 (include 6 and 21 in the number) (one 2 and one 3 and one 7)
LCM≈LCD
so
17.
denomenators are 2 and x^2
2=2
x^2=x times x
lcm=2 times x times x=2x^2
18.
denomenators are 6 and 9
6=2 and 3
9=3 and 3
LCM=2 times 3 times 3=18
19.
denomenators are z and 7z
z=z
7z=7 and z
lcm=7 times z=7z
20.
denomenators are 5b and 7b^3c
5b=5 and b
7b^3c=7 times b times b times b times c
LCM=7 times 5 times b times b times b times c=7b^3c=7cb^3
21. denomenator=5 and x+2
5=5
x+2=x+2
LCM=5 times (x+2)=5(x+2)=5x+10
22. denomenators are ab and b^2c
ab=a time b
b^2c=b times b times c
LCM=a times b times b times c=ab^2c=acb^2
23. demonenators are m+n and m-n
m+n=m+n
m-n=m-n
LCM=(m+n)(m-n)=m^2-n^2
24.denoenators are k and k^2-2
k=k
k^2-2=k^2-2
LCM=k times (k^2-2)=k^3-2k
Answer: No, it is not sufficient to serve 65 vehicles as it can only serve 62 vehicles with this space
Step-by-step explanation:
Since we have given that
Dimensions of space where vehicles are stored is
145 feet by 130 feet
Number of vehicles to serve = 65
Area required for each vehicle = 300 ft²
So, we need to find that is it sufficient for to serve 65 vehicles.
So, first we find the number of vehicles can be served in this space
No, it is not sufficient to serve 65 vehicles as it can only serve 62 vehicles with this space.
