All categories

3 years ago

You have cards with the letters A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P. Event U is choosing the cards A, B, C

Mathematics

1 answer:

Thepotemich [5.8K]3 years ago

5 0

Answer:

If you are asking how many events use P: 1 event is using the letter P (event W)

Step-by-step explanation:

Why event U cannot be using P: This event is only using the letters A, B, C, and D.

Why event V cannot be using P: This event is only using vowels (A, E, I, O, U)

You might be interested in

mike observed that 75% of the students of a school liked skating. if 35 students of the school didn't like skating, the number o

bezimeni [28]

Cross multiply.

75 / x = 25 / 35
25x = 35(75)
25x = 2625
/25 /25
x = 105

Therefore 105 students liked skating.

Proof:

105 + 35 = 140.
140 * 0.75 = 105
140 * 0.25 = 35

140 (total) split into quarters.

35 + 35 + 35 + 35 = 140
35 + 35 + 35 = 105 (3 quarters liked)
35 = 35 (1 quarter disliked)

3 0

3 years ago

Mae sells newspaper subscriptions door-to-door. In addition to a fixed salary, she earns a commission for each subscription that

Lady_Fox [76]

<span>The answer is y = 10x + 125</span>

6 0

3 years ago

Read 2 more answers

Can someone help me?

tatyana61 [14]

Answer:

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Considering only the values of β for which sinβtanβsecβcotβ is defined, which of the following expressions is equivalent to sinβ

-Dominant- [34]

Answer:

$\tan(\beta)$

Step-by-step explanation:

For many of these identities, it is helpful to convert everything to sine and cosine, see what cancels, and then work to build out to something. If you have options that you're building toward, aim toward one of them.

${\tan(\theta)}={\dfrac{\sin(\theta)}{\cos(\theta)}$ and ${\sec(\theta)}={\dfrac{1}{\cos(\theta)}$

Recall the following reciprocal identity:

$\cot(\theta)=\dfrac{1}{\tan(\theta)}=\dfrac{1}{ \left ( \dfrac{\sin(\theta)}{\cos(\theta)} \right )} =\dfrac{\cos(\theta)}{\sin(\theta)}$

So, the original expression can be written in terms of only sines and cosines:

$\sin(\beta)\tan(\beta)\sec(\beta)\cot(\beta)$

$\sin(\beta) * \dfrac{\sin(\beta) }{\cos(\beta) } * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) } {\sin(\beta) }$

$\sin(\beta) * \dfrac{\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} {\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}$

$\sin(\beta) *\dfrac{1 }{\cos(\beta) }$

$\dfrac{\sin(\beta)}{\cos(\beta) }$

Working toward one of the answers provided, this is the tangent function.

The one caveat is that the original expression also was undefined for values of beta that caused the sine function to be zero, whereas this simplified function is only undefined for values of beta where the cosine is equal to zero. However, the questions states that we are only considering values for which the original expression is defined, so, excluding those values of beta, the original expression is equivalent to $\tan(\beta)$ .

8 0

2 years ago

Help........... with this look

julsineya [31]

Hmmmmmm ok i know how to do this

3 0

2 years ago