All categories

3 years ago

Can anyone help me with this?

Mathematics

2 answers:

tangare [24]3 years ago

7 0

Answer:

around 12.08

Step-by-step explanation:

to do this, you need to do sine cosine tangent. Make sure your calculator is in degrees, not radians. Since the only other side given is the hypotenuse, 16, and y is adjacent to the angle, you use cosine (adjacent/hypotenuse). You do cos(41) and multiply it by 16 to get 12.08

dybincka [34]3 years ago

3 0

Answer:

12.1

Step-by-step explanation:

41 + x = 90

x= 49

Sine rule:

Sin90 = Sin49

16 y

y= 16sin49

y is approx. 12.1

You might be interested in

I-Ready sometimes gets me stumped.

Sonja [21]

Answer: D

Step-by-step explanation: it’s the last one, I think .

8 0

2 years ago

Read 2 more answers

Work out the area of the semi circle pi = 3.142 and radius 5

neonofarm [45]

Answer:

39.25

Step-by-step explanation:

Area of circle is pi(3.14) times r^2.

Semi means half

So 3.14 times 5^2 = 78.5

Its a semi ciricle so its 78.5/2 which is 39.25

8 0

2 years ago

Please helpppp, 10 points!

Jet001 [13]

This math problem is about rounding. So you need to round the probability of it for example there is a big probability that it will land on red again.

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

Find the vertex for the function: y = 6(x + 3)^2 - 1

sukhopar [10]

Answer:

The vertex would be (-3, -1)

Step-by-step explanation:

Since this is in vertex form already, a(x-h)^2+k where h is the x value of the vetex, and k is y.

We would take the opposite of +3 and get -3, which gives us our x cordinate.

We don't change the k value, so it just stays -1, giving us our y value.

Therefore, the vertex is at (-3, -1)

Hope this helps!

6 0

2 years ago