All categories

3 years ago

I'm not sure how to set this up

Mathematics

1 answer:

masya89 [10]3 years ago

5 0

So hmmm notice the picture below

recall the "inscribed angle" theorem, an angle inscribed is half the intercepted arc

now, the line BA goes through the center, and to each end, thus is a diameter, the point of tangency, where a tangent hits the diameter or radius of a circle, is always a right-angle

You might be interested in

363 · 62 . 46-4<br> Plsss help solve it :)

Andreas93 [3]

Answer:

1035272

Step-by-step explanation:

4 0

2 years ago

Equivalent fractions to 28/32

Scrat [10]

7/8 there you go the right answer xD

7 0

3 years ago

Read 2 more answers

Which fraction has a value greater than 0.4

hodyreva [135]

U can see that 5/9 has a greater value thx hope it helped

6 0

2 years ago

Read 2 more answers

I WILL MARK YOU THE BRAINLIEST!!!!!!!!! AND GIVE LOT'S OF POINTS On a road trip, a family drives 350 miles per day. How many day

UNO [17]

The family needs 4 days to cover 1,400 miles

4 0

2 years ago

A vector with magnitude 9 points in a direction 190 degrees counterclockwise from the positive x axis. Write in component form

Over [174]

Answer:

$\vec{v}= \text{ or } \approx$

Step-by-step explanation:

Component form of a vector is given by $\vec{v}=$ , where $i$ represents change in x-value and $j$ represents change in y-value. The magnitude of a vector is correlated the Pythagorean Theorem. For vector $\vec{v}=$ , the magnitude is $||v||=\sqrt{i^2+j^2$ .

190 degrees counterclockwise from the positive x-axis is 10 degrees below the negative x-axis. We can then draw a right triangle 10 degrees below the horizontal with one leg being $i$ , one leg being $j$ , and the hypotenuse of the triangle being the magnitude of the vector, which is given as 9.

In any right triangle, the sine/sin of an angle is equal to its opposite side divided by the hypotenuse, or longest side, of the triangle.

Therefore, we have:

$\sin 10^{\circ}=\frac{j}{9},\\j=9\sin 10^{\circ}$

To find the other leg, $i$ , we can also use basic trigonometry for a right triangle. In right triangles only, the cosine/cos of an angle is equal to its adjacent side divided by the hypotenuse of the triangle. We get:

$\cos 10^{\circ}=\frac{i}{9},\\i=9\cos 10^{\circ}$

Verify that $(9\sin 10^{\circ})^2+(9\cos 10^{\circ})^2=9^2\:\checkmark$

Therefore, the component form of this vector is $\vec{v}=\boxed{}\approx \boxed{}$

6 0

2 years ago