So for question one (about the equilateral triangle), you know any triangle equals 180 degrees. So an equilateral triangle has all equal angles, so 180/3 would equal 60 degrees for each angle.

Now for the first triangle out of the three isosceles.

You should know that there are two equal sides and one side that is not equal in a isosceles triangle.

So if one side is 34 degrees, and the other two are equal, then you first subtract 34 degrees from 180 degrees, because again, all triangles have 180 degrees. So 180-34 would equal to 146 degrees left. Now you divide 146 by two, because the other two angles are equal to each other. So 146/2 = 73 degrees for each of the remaining angles.

For the second triangle out of the three isosceles.

First, let's just calculate half of the triangle first. You already know two angles, 30 and 90 degrees (hence the 90 degree symbol).

So 30 + 90 would equal 120 degrees in total out of 180 degrees (because 1 triangle = 180 degrees).

So if you subtract 180-120 or 120 + ? = 180 to know the remaining angle of the half of the triangle, you would get 60 degrees left. And that also goes with the other side of the top angle of the entire triangle, so your answer for the second one would be 60 degrees. :-)

For the third triangle out of the three isosceles.

You know one angle is 75 degrees.

If you turn your screen a little bit (like about 90 degrees) to the left, you would see that two sides (with the x degree in the middle) is equal. So then know you know the other side of the triangle would also be 75 degrees, because one side that is equal has 75 degrees.

Now you have two 75 degrees, you add 75 + 75, which gives you 150 degrees now.

So obviously again, 180 degrees is the total amount of degrees in a triangle, and you want to know the last degree that would make the triangle equal to 180 degrees out of 150 degrees, so 180-150 or 150 + ? = 180, which gives you 30 degrees for x for the third isosceles triangle.

try doing the fourth one by yourself lol

5 0

3 years ago

HELP I dont understand-

Naddik [55]

Step-by-step explanation:

If a radius of a circle and a tangent to a circle intersect at the point of tangency, then the radius and the tangent are perpendicular.

Answer: B

8 0

2 years ago

Read 2 more answers

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).