All categories

3 years ago

Find Theta in degrees. WILL MARK BRAINLIEST I PROMISE<3333

Mathematics

1 answer:

Semmy [17]3 years ago

3 0

Answer:

$sin \alpha = \frac{opposite}{hypotenuse} \\ sin \alpha = \frac{8}{17} \\ sin \alpha = 0.470588235294117 \\ \alpha = {sin}^{ - 1} (0.470588235294117) \\ \alpha = 28.07 \: degrees$

You might be interested in

A particle travels along the x-axis such that its position at time t is given by the function x(t)= 2t^2+t. What is the average

Lapatulllka [165]

Answer:

4: A

Step-by-step explanation:

Mathematically, the average speed over a particular range will be;

f’(b)-f’(a)/(a-b)

In this case; a = 2 and b = 10

So we have the following;

f’(x) = 4t + 1

f’(10) = 4(10) + 1 = 41

f’(2) = 4(2) + 1 = 9

So we have the average speed change as;

(41-9)/(10-2)

= 32/8 = 4

8 0

3 years ago

URGENT!! 25 points. help on C please. this assignment is already late

andrew-mc [135]

Answer:

$7+(11-4\times 2)=10$

Step-by-step explanation:

So, this question involves you to play around with the numbers and the operations a bit. I've come up with a solution, but perhaps there are others as well.

So, we have:

$7\text{ } 11\text{ } 4\text{ } 2\text{ } =\text{ } 10$

So, what we can do is, for instance, multiply 4 and 2. Subtract that into 11. And then add the difference to 7. Thus:

$7+(11-4\times 2)=10$

This is a true number sentence. Let's do it to make sure.

First, do the operations inside the parentheses.

Multiplication comes first, so multiply:

$7+(11-8)=10$

Again, do the parentheses first. Subtract:

$7+(3)=10$

Now add:

$10=10$

So our answer is correct :)

5 0

3 years ago

Find the solution of the given initial value problem in explicit form. y′=(1−5x)y2, y(0)=−12 Enclose numerators and denominators

Nata [24]

Answer:

The solution is $y=-\frac{12}{12x-30x^2+1}$ .

Step-by-step explanation:

A first order differential equation $y'=f(x,y)$ is called a separable equation if the function $f(x,y)$ can be factored into the product of two functions of $x$ and $y$ :

$f(x,y)=p(x)h(y)$

where $p(x)$ and $h(y)$ are continuous functions.

We have the following differential equation

$y'=(1-5x)y^2, \quad y(0)=-12$

In the given case $p(x)=1-5x$ and $h(y)=y^2$ .

We divide the equation by $h(y)$ and move $dx$ to the right side:

$\frac{1}{y^2}dy\:=(1-5x)dx$

Next, integrate both sides:

$\int \frac{1}{y^2}dy\:=\int(1-5x)dx\\\\-\frac{1}{y}=x-\frac{5x^2}{2}+C$

Now, we solve for $y$

$-\frac{1}{y}=x-\frac{5x^2}{2}+C\\-\frac{1}{y}\cdot \:2y=x\cdot \:2y-\frac{5x^2}{2}\cdot \:2y+C\cdot \:2y\\-2=2yx-5yx^2+2Cy\\y\left(2x-5x^2+2C\right)=-2\\\\y=-\frac{2}{2x-5x^2+2C}$