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Irina18 [472]
4 years ago
11

Which point represents the intersection of the given lines? AE and DE

Mathematics
1 answer:
zalisa [80]4 years ago
6 0
Need a picture of what you asking 

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tabita bought peppers that cost 0.79 per pound she paid 3.95 for the peppers how many pounds did she buy
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237


Step-by-step explanation:


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Pravat used parking cones to set up a rectangular football field in the park. However, when his friend Brian arrived at the park
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C

Step-by-step explanation:

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4 years ago
If R is midpoint of QS and QS=40 find QR
Elodia [21]

Answer:

Given: R is the midpoint of QS.

QS=40

Now,

QR=40/2(R IS THE MIDPOINT)

=20

8 0
4 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
Write each fraction as a percent 2/7 roound desimal to nerest 100
ivolga24 [154]
20/7? I am guessing. Because that’s almost near the hundredth place
8 0
3 years ago
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