<h3>
Answer: ds/dt = 11</h3>
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Work Shown:
Before we can use derivatives, we need to find the value of s when (x,y) = (15,20)
s^2 = x^2+y^2
s^2 = 15^2+20^2
s^2 = 225+400
s^2 = 625
s = sqrt(625)
s = 25
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Now we can apply the derivative to both sides to get the following. Don't forget to use the chain rule.
s^2 = x^2 + y^2
d/dt[s^2] = d/dt[x^2 + y^2]
d/dt[s^2] = d/dt[x^2] + d/dt[y^2]
2s*ds/dt = 2x*dx/dt + 2y*dy/dt
2(25)*ds/dt = 2(15)*5 + 2(20)*(10)
50*ds/dt = 150 + 400
50*ds/dt = 550
ds/dt = 550/50
ds/dt = 11
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Side note: The information t = 40 is never used. It's just extra info.
Answer:
-1<x<9
(I think you notated it wrong, it should be |x-4|-2<3)
Ok so first we find the equation that equals one variable.
2y = -x + 9
3x - 6y = -15
We solve for y.
2y = -x + 9
y = -x/2 + 9/2
Then we plug in this y value into the other equation to keep only one variable so we can solve for it.
3x - 6y = -15
3(-x + 9/2) - 6y = -15
-3x + 27/2 - 6y = -15
-9y + 27/2 = -15
-9y = 3/2
-y = 3/18
y = -3/18
Then we plug in this numerical y-value into the first equation which we found out by solving an equation for y.
y = -x/2 + 9/2
-3/18 = -x/2 + 9/2
-84/18 = -x/2
-x = 9 1/3
x = -28/3
Your answer would be (-28/3, -3/18)
Hope this helps!
Answer:
m∠ADC = 132°
Step-by-step explanation:
From the figure attached,
By applying sine rule in ΔABD,
sin(∠ADB) = 
= 0.74231
m∠ADB = 
= 47.92°
≈ 48°
m∠ADC + m∠ADB = 180° [Linear pair of angles]
m∠ADC + 48° = 180°
m∠ADC = 180° - 48°
m∠ADC = 132°
