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3 years ago

9

Help me with this plz

1 answer:

djyliett [7]3 years ago

7 0

Ok what do u need help with?

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Number 10 <br> I think it's 1 but I'm not exactly sure

Tasya [4]

Hello,
answer is C

m=-40h+130
Indeed
A=(1/2,110)
B=(3,10)

y-10=(10-110/(3-1/2)(x-3)
==>y=-100/(5/2)*(x-3) +10
==>y=-40x+120+10
==>y=-40x+130
(or m=-40h+130 )

8 0

4 years ago

Square root (2x)^8 simplified

dangina [55]

Answer:

256x^8

Step-by-step explanation:

8 0

3 years ago

Somebody please help meeeee

azamat

Step-by-step explanation:

$\frac{2 + 3i}{4 - 5i}$

$\frac{(2 + 3i)(4 + 5i)}{(4 - 5i)(4 + 5i)}$

$\frac{8 + 10i + 12i + 15 {i}^{2} }{16 - 25 {i}^{2} }$

$\frac{8 + 22i + 15( -1 )}{16 - 25 ( - 1)}$

$\frac{8 - 15 + 22i}{16 + 25}$

$\frac{ - 7 + 22i}{41}$

So, a= -7

b= 22

c= 41

8 0

3 years ago

Use the graph of the function. Determine Over what interval(s) the function is positive or negative.

soldier1979 [14.2K]

Answer:

please where is the drawing I mean the graph

6 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