x = longer section
y = shorter section
x + y = 109
x = 2y + 10
so we sub in eq 2 into eq 1
2y + 10 + y = 109
3y + 10 = 109
3y = 109 - 10
3y = 99
y = 99/3
y = 33 <=== shorter section is 33 cm
x = 2y + 10
x = 2(33) + 10
x = 66 + 10
x = 76 <=== longer section is 76 cm
Answer:
(7,-11)
Step-by-step explanation:
We are given the rule of:
.
Apply the rule to point P:

P' should be (7,-11).
Hope this helps.
Answer:
Step-by-step explanation:
The max and min values exist where the derivative of the function is equal to 0. So we find the derivative:

Setting this equal to 0 and solving for x gives you the 2 values
x = .352 and -3.464
Now we need to find where the function is increasing and decreasing. I teach ,my students to make a table. The interval "starts" at negative infinity and goes up to positive infinity. So the intervals are
-∞ < x < -3.464 -3.464 < x < .352 .352 < x < ∞
Now choose any value within the interval and evaluate the derivative at that value. I chose -5 for the first test number, 0 for the second, and 1 for the third. Evaluating the derivative at -5 gives you a positive number, so the function is increasing from negative infinity to -3.464. Evaluating the derivative at 0 gives you a negative number, so the function is decreasing from -3.464 to .352. Evaluating the derivative at 1 gives you a positive number, so the function is increasing from .352 to positive infinity. That means that there is a min at the x value of .352. I guess we could round that to the tenths place and use .4 as our x value. Plug .4 into the function to get the y value at the min point.
f(.4) = -48.0
So the relative min of the function is located at (.4, -48.0)
The solutions for the following expressions are:
1. 0.5*2 = 1
2. -0.2+-0.05= -0.25
3. absolute value of (-3+2.75) = 0.25
4. -(-0.25) = 0.25
<span>5. 2x-1.75x = 0.25x</span><span>
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