Answer:
C. (1, –1); minimum
Step-by-step explanation:
The answer to your question is letter “C”
The correct answer is c
Answer:
m<1 = 39
m<2 = 51
Step-by-step explanation:
For this problem, you need to understand that a little square in the bottom of two connecting lines represents a right-angle (an angle this 90 degrees). This problem, gives you two relationships for angle 1 and angle 2 within a right-angle. Using this information, we can solve for the measures of the two angles.
Let's write the two relations:
m< 1 = 3x
m< 2 = x + 38
And now let's right an equation that represents the two angles to the picture:
m<1 + m<2 = 90
Using this information, let's substitute the expressions we have for the two angles and solve for x. Once we have the value of x, we can find the measure of the two angles.
m< 1 + m< 2 = 90
(3x) + (x + 38) = 90
3x + x + 38 = 90
x ( 3 + 1 ) + 38 = 90
x ( 4 ) + 38 = 90
4x + 38 = 90
4x + 38 - 38 = 90 - 38
4x = 90 - 38
4x = 52
4x * (1/4) = 52 * (1/4)
x = 52 * (1/4)
x = 13
Now that we have the value of x, we simply plug it back into our expressions for the m<1 and m<2.
m<1 = 3x = 3(13) = 39
m<2 = x + 38 = 13 + 38 = 51
And we can verify this is correct with the relational equation:
m<1 + m<2 = 90
39 + 51 ?= 90
90 == 90
Hence, we have found the values of m<1 and m<2.
Cheers.
This is so provided that the velocity changes continuously in which case we can apply the mean value theorem.
<span>Velocity (v) is the derivative of displacement (x) : </span>
<span>v = dx/dt </span>
<span>Monk 1 arrives after a time t* and Monk 2 too. </span>
<span>Name v1(t) and v2(t) their respective velocities throughout the trajectory. </span>
<span>Then we know that both average velocities were equal : </span>
<span>avg1 = avg2 </span>
<span>and avg = integral ( v(t) , t:0->t*) / t* </span>
<span>so </span>
<span>integral (v1(t), t:0->t*) = integral (v2(t), t:0->t*) </span>
<span>which is the same of saying that the covered distances after t* seconds are the same </span>
<span>=> integral (v1(t) - v2(t) , t:0->t*) = 0 </span>
<span>Thus, name v#(t) = v1(t) - v2(t) , then we obtain </span>
<span>=> integral ( v#(t) , t:0->t*) = 0 </span>
<span>Name the analytical integral of v#(t) = V(t) , then we have </span>
<span>=> V(t*) - V(0) = 0 </span>
<span>=> V(t*) = V(0) </span>
<span>So there exist a c in [0, t*] so that </span>
<span>V'(c) = (V(t*) - V(0)) / (t* - 0) (mean value theorem) </span>
<span>We know that V(0) = V(t*) = 0 (covered distances equal at the start and finish), so we get </span>
<span>V'(c) = v#(c) = v1(c) - v2(c) = 0 </span>
<span>=> v1(c) = v2(c) </span>
<span>So there exist a point c in [0, t*] so that the velocity of monk 1 equals that of monk 2. </span>
Use this to help you. this is a chart. it is not the answer but this helps
The answer is -1.01
I first changed the signs of the ones i could. So <span>-0.75+(2/50)+0.4(-3/4)
Then i multiplied the 0.4*-3/4 which equals -0.3
So you are left with </span><span>-0.75+(2/50)-0.3
2/50 is equal to 0.04 so now it is </span> -0.75+0.04-0.3 and the rest is simple math. Which equals -1.01
