Answer: 25% of 80 = 20
Step-by-step explanation:
(25:100)*80 =
(25*80):100 =
2000:100 = 20
- R3KTFORGOOD ☕
Answer:
x= 20
Step-by-step explanation:
Solve for x:
(7 x)/10 = 14
Multiply both sides of (7 x)/10 = 14 by 10/7:
(10×7 x)/(7×10) = 10/7×14
10/7×7/10 = (10×7)/(7×10):
(10×7)/(7×10) x = 10/7×14
10/7×14 = (10×14)/7:
(10×7 x)/(7×10) = (10×14)/7
(10×7 x)/(7×10) = (7×10)/(7×10)×x = x:
x = (10×14)/7
14/7 = (7×2)/7 = 2:
x = 10×2
10×2 = 20:
Answer: x = 20
Answer:
x is equal to 7
Step-by-step explanation:
Because the two horizontal lines are parallel, we know that the angles at which they intersect the diagonal line will be identical.
We also know that on either side of the diagonal line, the sum of the angles between that and either of the horizontal lines will be 180.
In short, we know that the two values given have a sum of 180. With that we can quickly solve for x:
(x + 67) + (x + 127) = 180
2x + 194 = 180
2x = -14
x = 7
Answer:
294 dollars
Step-by-step explanation:
find out how much 1 ticket cost by driving 6 from 126
then take that answer and multiply it times 14 and that's your answer
Here is one way to show that the left side is identical to the right side.

Throughout the entire process, the right hand side stayed the same. When working on identities, it's important to keep one side the same while you transform the other side.
Despite the two sides being identical when simplified, there is the issue of cos(x) being zero in the denominator on the left hand side. Be sure to account for this when forming the domain. No such issue occurs on the right hand side. For instance, the value x = pi/2 leads to a division by zero error when in radian mode. So if I was your teacher, I would revise the "for any value of x" and replace it with something along the lines of "for any x value in the domain".