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3 years ago

Please please help me

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

6 0

We can’t help u if their is no instructions on what u need to do sorry

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Mis himley picked 46 tomatoes from her garden on friday. On saturday she picked 17 tomatoes. How many tomatoes did she pick?

Leya [2.2K]

46 + 17 = 63. totally tomatoes she picked

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3 years ago

160 degrees 20 degrees 160 degrees 20 degrees name the shape

Gennadij [26K]

Answer: A rhombus

Step-by-step explanation:

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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

School x and y are in a comepetition. School x has 8 more points than school y and 3 times more points as school y . How many po

lara [203]

Answer: School Y has 4 points and School X has 12 points.

Explanation:

Let the number of points School Y has be x

Let the number of points School X has be 3x

According to question, we get

$3x-x=8\\\\2x=8\\\\x=\frac{8}{2}\\\\x=4$

So, the number of point School Y has 4.

and number of points School X has

$3x=3\times 4=12$

Hence, School Y has 4 points and School X has 12 points.

6 0

3 years ago

Use the equation p=b-3 to find the value of p when b = 8

Nataliya [291]

Answer:

Step-by-step explanation:

p=(8)-3

p=5

6 0

3 years ago

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Please please help me​

Please please help me