Ask question

Ask question

All categories

2 years ago

11

In Diagram 6, JCK is a tangent to the circle ABC

1 answer:

Igoryamba2 years ago

5 0

Answer:

C

Step-by-step explanation:

if you don't know it choose C

You might be interested in

Does this graph below represent a function? (Yes or no)

Whitepunk [10]

Answer:yes

Step-by-step explanation: it represents a function

4 0

2 years ago

Use prime factorization to reduce the fraction 36/180

laila [671]

Answer:

Step-by-step explanation:

36 = 6* 6 = 2* 3 * 2* 3 = 2*2*3*3

180 = 2 * 2 * 3 * 3 * 5 = 2 * 2 * 3*3*5

$\frac{36}{180}= \frac{2*2*3*3}{2*2*3*3*5} = \frac{1}{5}$

Factors that are common will be cancelled

5 0

2 years ago

1. It costs $3.45 to buy 3/4 lb of chopped walnuts. How much would it cost to purchase

konstantin123 [22]

Answer:

$34.50

Step-by-step explanation:

First, how many 3/4 lbs are in 7.5 lbs?

7.5 lbs = 15/2 lbs.

(15/2) / (3/4) = 10

There are 10 "3/4" lbs in 7.5 lbs.

So, if it costs $3.45 to purchase 3/4 lbs of chopped walnuts, it would cost $3.45 times 10 dollars to purchase 7.5 lbs of chopped walnuts, or $34.50.

Let me know if this helps!

5 0

2 years ago

The sum of two number is 10. the difference between the numbers is 4. what are the two numbers.

dolphi86 [110]

Your answer is 7 and 3.

3 0

3 years ago

Prove by mathematical induction that

postnew [5]

For $n=1$ , on the left we have $\cos\theta$ , and on the right,

$\dfrac{\sin2\theta}{2\sin\theta}=\dfrac{2\sin\theta\cos\theta}{2\sin\theta}=\cos\theta$

(where we use the double angle identity: $\sin2\theta=2\sin\theta\cos\theta$ )

Suppose the relation holds for $n=k$ :

$\displaystyle\sum_{n=1}^k\cos(2n-1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}$

Then for $n=k+1$ , the left side is

$\displaystyle\sum_{n=1}^{k+1}\cos(2n-1)\theta=\sum_{n=1}^k\cos(2n-1)\theta+\cos(2k+1)\theta=\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta$

So we want to show that

$\dfrac{\sin2k\theta}{2\sin\theta}+\cos(2k+1)\theta=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

On the left side, we can combine the fractions:

$\dfrac{\sin2k\theta+2\sin\theta\cos(2k+1)\theta}{2\sin\theta}$

Recall that

$\cos(x+y)=\cos x\cos y-\sin x\sin y$

so that we can write

$\dfrac{\sin2k\theta+2\sin\theta(\cos2k\theta\cos\theta-\sin2k\theta\sin\theta)}{2\sin\theta}$

$=\dfrac{\sin2k\theta+\sin2\theta\cos2k\theta-2\sin2k\theta\sin^2\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta(1-2\sin^2\theta)+\sin2\theta\cos2k\theta}{2\sin\theta}$

$=\dfrac{\sin2k\theta\cos2\theta+\sin2\theta\cos2k\theta}{2\sin\theta}$

(another double angle identity: $\cos2\theta=\cos^2\theta-\sin^2\theta=1-2\sin^2\theta$ )

Then recall that

$\sin(x+y)=\sin x\cos y+\sin y\cos x$

which lets us consolidate the numerator to get what we wanted:

$=\dfrac{\sin(2k+2)\theta}{2\sin\theta}$

and the identity is established.

8 0

2 years ago