Solve the right triangle round your answer to the nearest tenth

What are the common multipls of 2 and 3 between 2 and 25

Colt1911 [192]

Answer:

6, 12, 18, and 24

Step-by-step explanation:

Multiples of 2: 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24. Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24. Common multiples of 2 and 3 include 6, 12, 18, and 24.

4 0

2 years ago

What is the percent of change from 6,000 to 60?

Thepotemich [5.8K]

Answer: 99% decrease

Step-by-step explanation: Since the number changes from 6,000 to 60, we know that it's decreasing. Now to find the percent decrease, we divide the amount of change by the original number.

The amount of change is the difference between 6,000 and 60 and the original number is 6,000.

6,000 - 60 is 5,940 so we are left with $\frac{5,940}{6,000}$ and dividing 6,000 into 5,940 gives us 0.99.

Finally, since the problem is asking us for the percent, we write 0.999 as a percent by moving the decimal point 2 places to the right to get 99%.

So when a number changes from 6,000 to 60, it has decreased by 99%.

6 0

2 years ago

What rational number is equal to -2/3

valkas [14]

Answer:

Here is a little prompt to help you out.

Step-by-step explanation:

Two rational numbers are said to be equivalent if we divide each decimal value will be the same. that is if a/b is equivalent to c/d if a/b=p.

7 0

2 years ago

2p + 8 = 20-P 2p + 8 + p = 20-p+p 3p + 8 = 20 what’s the value of p ?

adell [148]

The answer to p is 4

7 0

2 years ago

Read 2 more answers

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