Answer:
P = 321.3m
Step-by-step explanation:
To help simplify the process of solving this problem, first break it up into different shapes. Here, you have two shapes: a rectangle and a circle (1 semi-circle + 1 semi-circle = 1 circle).
Next, use the equation for circumference of a circle to find the first part of the perimeter (C = πd). In this case, the diameter would be 45m.
- C = πd
- C = 3.14 × 45
- C = 141.3 m
This is the first part of the circumference. Now, you must consider the rectangle. (Because the two shorter sides of the rectangle are not part of the perimeter, do not include these in your final answer.) The length of the rectangle is 90 meters, and since both sides that measure to 90m are part of what makes up the perimeter, lastly add these to the circumference you calculated earlier.
- (90 + 90) or (90 × 2) = 180
- P = 141.3 + 180
- P = 321.3m
Hope this helps! : )
3x + 2y = 12
subtract 3x from both sides
2y = -3x + 12
divide all terms by 2 so that you can have only the y on the left side
y = -3/2x + 6
And that is your answer!
Hope this helped!! :)
Answer:
35 ia the answer sorry if im wrong
Please, use parentheses to enclose each fraction:
y=3/4X+5 should be written as <span>y=(3/4)X+5
Let's eliminate the fraction 3/4 by multiplying the above equation through by 4:
4[y] = 4[(3/4)x + 5]
Then 4y = 3x + 20
(no fraction here)
Let 's now solve the system
4y=3x + 20
4x-3y=-1
We are to solve this system using subtraction. To accomplish this, multiply the first equation by 3 and the second equation by 4. Here's what happens:
12y = 9x + 60 (first equation)
16x-12y = -4, or -12y = -4 - 16x (second equation)
Then we have
12y = 9x + 60
-12y =-16x - 4
If we add here, 12y-12y becomes zero and we then have 0 = -7x + 56.
Solving this for x: 7x = 56; x=8
We were given equations
</span><span>y=3/4X+5
4x-3y=-1
We can subst. x=8 into either of these eqn's to find y. Let's try the first one:
y = (3/4)(8)+5 = 6+5=11
Then x=8 and y=11.
You should check this result. Subst. x=8 and y=11 into the second given equation. Is this equation now true?</span>
