2p Determine the value of a such that the system of equations below would have an infinite number of solutions?
18x + 12y = 36
ax – 8y =- 24
Answer: a=-12
I'll do the first two to get you started
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Problem 1
A = 3 = starting value
B = 10 = ending value
C = percent change
C = [ (B - A)/A ] * 100%
C = [ (10-3)/3 ] * 100%
C = (7/3) * 100%
C = 2.3333333 * 100%
C = 233.33333%
C = 233.3%
The positive C value means we have a percent increase. If C was negative, then we'd have a percent decrease.
<h3>Answer: 233.3% increase</h3>
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Problem 2
A = 9 = start value
B = 20 = end value
C = percent change
C = [ (B - A)/A ] * 100%
C = [ (20-9)/9 ] * 100%
C = (11/9)*100%
C = 1.2222222222*100%
C = 122.22222222%
C = 122.2%
<h3>Answer: 122.2% increase</h3>
Answer:
x= 5
Step-by-step explanation:
both lines are the same length
5x-6=2x+9
-2x -2x
3x-6= +9
+6 +6
3x = 15
÷3 ÷3
x= 5
Answer:
D. 
Step-by-step explanation:
There is a translation 1 point up along the y axis and a compression of 4.
Moving a function up (let's use <em>h</em> for the amount of points up) would change the function as so:
Meanwhile, the compression would modify x in this case. You can eliminate any answers (A. and B.) that have no modification to x, and eliminate C., as a fraction modification would actually widen the graph instead of compress it.
Hope this helps! :]
Answer:
C(p) = 4,96 (in thousands of dollars)
l = 2980 $ invest in labor
k = 2980 $ invest in equipment
Step-by-step explanation:
Information we have:
Monthly output P = 450*l*k ⇒ k = P/450*l
But the production need to be 4000
Then k = 4000/450*l
Cost of production = l * k (in thousands of dollars)
C(l) = l + 4000/450*l
Taking derivatives (both members of the equation)
C´(l) = 1 - 400 /45*l² ⇒ C´(l) = 0 ⇒ 1 - 400/45l² = 0
45*l² - 400 = 0 ⇒ l² = 400/45
l = 2.98 (in thousands of dollars)
l = 2980 $ And
k = 400/45*l ⇒ k 400/45*2.98
k = 2.98 (in thousands of dollars)
C(p) = l + k
C(p) = 2980 + 2980
C(p) = 5960 $
