Let's begin by calling Sarah's age now as X. As Ralph is 3 times as old as Sarah, X times 3 = 3X. Hence, Ralph's age is 3X. In six years, Ralph will be twice as old as Sarah. To calculate six years from now, add 6 to X for Sarah, and 6 to 3X for Ralph. As Ralph is twice as old as Sarah and we want to find the difference between the ages to calculate X, multiply X+6 by 2. You'll get 2X+12. Therefore, 2X+12=3X+6. Deduct 6 from 3X+6 as we want to isolate the variable. Because you did that to one side, you have to deduct 6 from 2X+12. Hence, now you have 2X+6=3X. X=6. Ralph's age is 3X, so 6 times 3 is 18. Ralph is 18 years old.
Answer:
Option A is correct
Step-by-step explanation:
Please Mark me brainlist
A. 2 parts is 7, so 7/2 is 3.5 = 1 part, then multiply by 3 to make the corresponding side which is 10.5 or 10 1/2
−10x+5y=10
x−5y=−28
Rewrite equations:
x−5y=−28
−10x+5y=10
Step: Solve:
x−5y=−28
x−5y+5y=−28+5y(Add 5y to both sides)
x=5y−28
Step: Substitute:
−10x+5y=10
−10(5y−28)+5y=10
−45y+280=10(Simplify both sides of the equation)
−45y+280+−280=10+−280(Add -280 to both sides)
−45y=−270
−45y
−45
=
−270
−45
(Divide both sides by -45)
y=6
Step: Substitute:
x=5y−28
x=(5)(6)−28
x=2(Simplify both sides of the equation)
Answer:
x=2 and y=6
The question is incomplete, as the depth of the hook isn't given in the question. However, using a proposed depth for the hook, we could give a stepwise solution to the exercise.
Answer:
Kindly check explanation
Step-by-step explanation:
If given a hook depth of 6 meters, and a distance of 8 m(fishing line from hook) at the the Same depth, then the distance between rose and the fish, x can be calculated, using the relation.
Hypotenus² = opposite² + adjacent²
(refer to attached picture)
x² = 6² + 8²
x² = 36 +. 64
x² = 100
x = sqrt(100)
x = 10 meters
You can use the actual depth value to obtain the actual solution to your question.