what are the coordinates of a point on the unit circle if the angle formed by the positive x-axis and the radius is 60

(5x10^3)x(9x10^7) can someone please answer this for me.

Anna11 [10]

4.5 times 10^10

1. Multiply 5times ten to the third

2. Multiply 9 times ten raised to the seventh

Then multiply them together
5000 times 90000000

5 0

3 years ago

In order to qualify for a role in a play, an actor must be taller than 64 inches but shorter than 68 inches. The inequality 64 &

netineya [11]

Answer: x > 64 and x < 68 is another way of rewriting inequality.

Step-by-step explanation:

To qualify for a role in a play, an actor must be taller than 64 inches but shorter than 68 inches.

⇒The inequality 64 < x < 68, where x represents height,gives the range of height.

⇒In other words range of height can be x > 64 and x < 68 , such that both conditions are applied to single actor.

[with the use of or it will appear to give priority by actor to follow any 1 conditon]

8 0

2 years ago

a store is giving away $10 coupons to every 7th person to enter the store and a $25 coupon to every 18th person too emter the st

almond37 [142]

7,14,21,28,35,42,49,56,63,70,77,84,91,98,105,112,119,126
18,36,54,72,90,108,126

So the first person to get both of them is the 126th person

3 0

3 years ago

Rectangle EFGH is similar to rectangle JKLM. Which proportion can be used to

Dominik [7]

*See attached picture for the diagrams being referred to

Answer:

C. ⁵/12= ²/x

Step-by-step Explanation:

Given that rectangle EFGH is similar to rectangle JKLM, it means the corresponding sides of both rectangles are similar, and as such the ratios if the corresponding sides of rectangle EFGH and rectangle JKLM would be equal.

Thus, EF/JK = FG/KL

Since, EF = 5; JK = 12; FG = 2; and KL= x, therefore, the proportion to use in finding x would be:

⁵/12= ²/x

6 0

3 years ago

An urn contains n white balls andm black balls. (m and n are both positive numbers.) (a) If two balls are drawn without replacem

Genrish500 [490]

DISCLAIMER: Please let me rename b and w the number of black and white balls, for the sake of readability. You can switch the variable names at any time and the ideas won't change a bit!

Case 1: both balls are white.

At the beginning we have $b+w$ balls. We want to pick a white one, so we have a probability of $\frac{w}{b+w}$ of picking a white one.

If this happens, we're left with $w-1$ white balls and still $b$ black balls, for a total of $b+w-1$ balls. So, now, the probability of picking a white ball is

$\dfrac{w-1}{b+w-1}$

The probability of the two events happening one after the other is the product of the probabilities, so you pick two whites with probability

$\dfrac{w}{b+w}\cdot \dfrac{w-1}{b+w-1}=\dfrac{w(w-1)}{(b+w)(b+w-1)}$

Case 2: both balls are black

The exact same logic leads to a probability of

$\dfrac{b}{b+w}\cdot \dfrac{b-1}{b+w-1}=\dfrac{b(b-1)}{(b+w)(b+w-1)}$

These two events are mutually exclusive (we either pick two whites or two blacks!), so the total probability of picking two balls of the same colour is

$\dfrac{w(w-1)}{(b+w)(b+w-1)}+\dfrac{b(b-1)}{(b+w)(b+w-1)}=\dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Case 1: both balls are white.

In this case, nothing changes between the two picks. So, you have a probability of $\frac{w}{b+w}$ of picking a white ball with the first pick, and the same probability of picking a white ball with the second pick. Similarly, you have a probability $\frac{b}{b+w}$ of picking a black ball with both picks.

This leads to an overall probability of

$\left(\dfrac{w}{b+w}\right)^2+\left(\dfrac{b}{b+w}\right)^2 = \dfrac{w^2+b^2}{(b+w)^2}$

Of picking two balls of the same colour.

We want to prove that

$\dfrac{w^2+b^2}{(b+w)^2}\geq \dfrac{w(w-1)+b(b-1)}{(b+w)(b+w-1)}$

Expading all squares and products, this translates to

$\dfrac{w^2+b^2}{b^2+2bw+w^2}\geq \dfrac{w^2+b^2-b-w}{b^2+2bw+w^2-b-w}$

As you can see, this inequality comes in the form

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k}$

With x and y greater than k. This inequality is true whenever the numerator is smaller than the denominator:

$\dfrac{x}{y}\geq \dfrac{x-k}{y-k} \iff xy-kx \geq xy-ky \iff -kx\geq -ky \iff x\leq y$

And this is our case, because in our case we have

$x=b^2+w^2$
$y=b^2+w^2+2bw$ so, y has an extra piece and it is larger
$k=b+w$ which ensures that k<x (and thus k<y), because b and w are integers, and so b<b^2 and w<w^2

4 0

3 years ago