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dalvyx [7]
2 years ago
14

2.) Mila completed a half marathon race over the weekend. The full race was

Mathematics
1 answer:
In-s [12.5K]2 years ago
7 0

Answer:

10.48 miles.

Step-by-step explanation:

That is 4/5 * 13.1

= 10.48 miles.

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It should be Letter B
8 0
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Can you help me with this question please
solniwko [45]

Answer:

the answer is 9^8

Step-by-step explanation:

9^3 * 9^5 = 9^ 3 + 5 = 9^8

7 0
3 years ago
Which is the value of the expression? (-2) to the fourth power
kompoz [17]

Answer:

Variant D

Step-by-step explanation:

-2*(-2)*(-2)*(-2)=16

5 0
3 years ago
Consider the curve of the form y(t) = ksin(bt2) . (a) Given that the first critical point of y(t) for positive t occurs at t = 1
mafiozo [28]

Answer:

(a).   y'(1)=0  and    y'(2) = 3

(b).  $y'(t)=kb2t\cos(bt^2)$

(c).  $ b = \frac{\pi}{2} \text{ and}\  k = \frac{3}{2\pi}$

Step-by-step explanation:

(a). Let the curve is,

$y(t)=k \sin (bt^2)$

So the stationary point or the critical point of the differential function of a single real variable , f(x) is the value x_{0}  which lies in the domain of f where the derivative is 0.

Therefore,  y'(1)=0

Also given that the derivative of the function y(t) is 3 at t = 2.

Therefore, y'(2) = 3.

(b).

Given function,    $y(t)=k \sin (bt^2)$

Differentiating the above equation with respect to x, we get

y'(t)=\frac{d}{dt}[k \sin (bt^2)]\\ y'(t)=k\frac{d}{dt}[\sin (bt^2)]

Applying chain rule,

y'(t)=k \cos (bt^2)(\frac{d}{dt}[bt^2])\\ y'(t)=k\cos(bt^2)(b2t)\\ y'(t)= kb2t\cos(bt^2)  

(c).

Finding the exact values of k and b.

As per the above parts in (a) and (b), the initial conditions are

y'(1) = 0 and y'(2) = 3

And the equations were

$y(t)=k \sin (bt^2)$

$y'(t)=kb2t\cos (bt^2)$

Now putting the initial conditions in the equation y'(1)=0

$kb2(1)\cos(b(1)^2)=0$

2kbcos(b) = 0

cos b = 0   (Since, k and b cannot be zero)

$b=\frac{\pi}{2}$

And

y'(2) = 3

$\therefore kb2(2)\cos [b(2)^2]=3$

$4kb\cos (4b)=3$

$4k(\frac{\pi}{2})\cos(\frac{4 \pi}{2})=3$

$2k\pi\cos 2 \pi=3$

2k\pi(1) = 3$  

$k=\frac{3}{2\pi}$

$\therefore b = \frac{\pi}{2} \text{ and}\  k = \frac{3}{2\pi}$

7 0
4 years ago
Solve the following system of equations for x to the nearest hundredth : y + 2x + 1 = 0; 4y - 4x ^ 2 - 12x = - 7
Sever21 [200]

Answer:

+3.464; -3.464

Step-by-step explanation:

call A = y + 2x + 1 = 0 => y = (1 - 2x)

call B: 4y - 4(x^2) - 12x = -7

=> replace y from A to B =>

  1. 4(1 - 2x) - 4(x^2) - 12x = -7
  2. 4 - 8x - 4(x ^ 2) - 12x = -7
  3. -8x - 4(x ^ 2) - 12x = -7 - 4 = -11
  4. -4(x^2) - (8x - 12x) = -11
  5. -4(x^2) + 4x = -11
  6. -4(x^2) + 4x + 11 = 0

=> get delta Δ = (-4^2) - 4*(-4 * 11) = 192

=> Δ > 0 => got 2 No

=> x1 = \frac{-4 + \sqrt{192} }{2 * -4} = \frac{1 - 2\sqrt{3} }{2} = -1.232

=> x2 = \frac{-4 - \sqrt{192} }{2 * -4}=\frac{1 + 2\sqrt{3} }{2}= 2.232

=> replace x from B into A

=> y1 = (1 - 2x) = (1 - 2 * -1.232) = 3.464

=> y2 = (1 - 2x) = (1 - 2 * 2.232) = - 3.464

7 0
3 years ago
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