You do 5 times x and 5 times 4 and you’d end up with equation that says 5x-4=6. The you add 4 to both sides of the equation to cancel out the 4. Your equation should now say 5x=10. You divide both sides by 5 and you have your answer x=2
Because these are on opposite sides of the parallelogram, x + 33 = 4x + 9.
This gives you x = 8; OL = 41, NM = 41.
Answer:
9/25x
Step-by-step explanation:
3/5/x ×12/20
3/5×1/x ×12/20
36/100x
Divide both by 4
9/25x
The exact value of x cannot be determined because we don't know what the equation is equated to
So we will have to leave the answer as 9/25x
The key is Esther travelled the same distance - x - in both her morning and evening commute.
45(time she took in the morning, or p) = x
30(time she took in the evening, or q) = x
Therefore 45(p) = 30(q), or divide both sides by 5 and get 9(p) = 6(q). I know you can divide it further, but these numbers are small enough and it's not worth the time.
Since the whole trip took an hour, (p + q) = 60min, and so, p = 60-q.
Therefore 9(60-q) = 6q or 540-9q = 6q. So 540 = 15q, which makes q = 36. If q = 36, then by (p+q)=60, p (the time she took in the morning) must equal 24.
45 miles per hour, her speed in the morning, times (24/60) hours, her time, makes 18 miles travelled in the morning. If you check, 30 miles per hour times (36/60) hours also makes 18 miles in the evening.
<span>Hope that makes a little sense. And I also hope it's right</span>
Answer:
-2, 8/3
Step-by-step explanation:
You can consider the area to be that of a trapezoid with parallel bases f(a) and f(4), and width (4-a). The area of that trapezoid is ...
A = (1/2)(f(a) +f(4))(4 -a)
= (1/2)((3a -1) +(3·4 -1))(4 -a)
= (1/2)(3a +10)(4 -a)
We want this area to be 12, so we can substitute that value for A and solve for "a".
12 = (1/2)(3a +10)(4 -a)
24 = (3a +10)(4 -a) = -3a² +2a +40
3a² -2a -16 = 0 . . . . . . subtract the right side
(3a -8)(a +2) = 0 . . . . . factor
Values of "a" that make these factors zero are ...
a = 8/3, a = -2
The values of "a" that make the area under the curve equal to 12 are -2 and 8/3.
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<em>Alternate solution</em>
The attachment shows a solution using the numerical integration function of a graphing calculator. The area under the curve of function f(x) on the interval [a, 4] is the integral of f(x) on that interval. Perhaps confusingly, we have called that area f(a). As we have seen above, the area is a quadratic function of "a". I find it convenient to use a calculator's functions to solve problems like this where possible.
