All categories

2 years ago

Which of the following are remote interior angles of angle 1? Check all that apply

Mathematics

2 answers:

Anni [7]2 years ago

8 0

Step-by-step explanation:

Ksenya-84 [330]2 years ago

5 0

Answer:

The answer would be 6 and 4.

Step-by-step explanation:

Because when they mean remote interior, they are hinting that the interior angles that are not close to 1. We cannot choose 3 in this situation because it is adjacent to angle 1.

You might be interested in

Anyone good at math? here's an easy one but im dum so please help will give brainliest

luda_lava [24]

Answer:

1.5 I'm pretty shure

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

I need help solving this inverse function. Does anybody know how to do it? I'm almost gonna be having my finals so I wanna be re

Sphinxa [80]

What is the question

3 0

2 years ago

What does this equation mean? 35= 5 x 7

Vlad1618 [11]

It mean something that is even

4 0

3 years ago

Read 2 more answers

Number 4 i don’t know how to do this one

Kruka [31]

Answer:

r= 3 h = 6 then use 3^2 + 6^2 = l^2

Step-by-step explanation:

3 0

2 years ago

Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

lakkis [162]

Answer:

$2*sin(x)+y*cos(x)-cos(y)=C_1$

Step-by-step explanation:

Let:

$P(x,y)=2*cos(x)-y*sin(x)$

$Q(x,y)=cos(x)+sin(y)$

This is an exact differential equation because:

$\frac{\partial P(x,y)}{\partial y} =-sin(x)$

$\frac{\partial Q(x,y)}{\partial x}=-sin(x)$

With this in mind let's define f(x,y) such that:

$\frac{\partial f(x,y)}{\partial x}=P(x,y)$

and

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate $\frac{\partial f(x,y)}{\partial x}$ with respect to x in order to find f(x,y)

$f(x,y)=\int\ 2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)$

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}$

Now, let's replace the previous result into $\frac{\partial f(x,y)}{\partial y}=Q(x,y)$ :

$cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)$

Solving for $\frac{dg(y)}{dy}$

$\frac{dg(y)}{dy}=sin(y)$

Integrating both sides with respect to y:

$g(y)=\int\ sin(y) \, dy =-cos(y)$

Replacing this result into f(x,y)

$f(x,y)=2*sin(x)+y*cos(x)-cos(y)$

Finally the solution is f(x,y)=C1 :

$2*sin(x)+y*cos(x)-cos(y)=C_1$

7 0

3 years ago