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3 years ago

14

Aubreys dinner cost $85 she tips the waitstaff 30% for excellent service

1 answer:

Pavel [41]3 years ago

4 0

Answer:

25.5

Step-by-step explanation:

You might be interested in

Which expression is equivalent to -16 - 7?

lakkis [162]

Step-by-step explanation:

B. -16 + (-7) ...this is the same as -16-7

3 0

3 years ago

Let M be the set of all nxn matrices. Define a relation on won M by A B there exists an invertible matrix P such that A = P BP S

Sophie [7]

Answer:

Recall that a relation is an <em>equivalence relation</em> if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

<em>Reflexive: </em>We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

<em>Symmetric</em>: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

<em>Transitive</em>: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get

$A=P^{-1}Q^{-1}CQP = (P^{-1}Q^{-1})C(QP)$ .

Recall that if P and Q are invertible, then QP is invertible and $(QP)^{-1}=P^{-1}Q^{-1}$ . So, if we denote R=QP we obtained that

$A=R^{-1}CR$ . Hence, A↔C.

Therefore, the relation is an <em>equivalence relation</em>.

4 0

3 years ago

Can someone please help me with this?

alisha [4.7K]

i cant read it

Answer:

j

Step-by-step explanation:

3 0

3 years ago

Rachel plays a game by rolling two number cubes with sides numbered 1 through 6 to win the game the sum of the numbers facing up

Artist 52 [7]

The probability that Rachel will win the game is: 1/12

Step-by-step explanation:

The number cubes has six sides numbered between 1 to 6. the chances of each number are equally likely

Let S be the sample space

The sample space has 6*6 = 36 outcomes.

Now, Let A be the event that the sum of numbers on both number cubes is 10

A = {(4,6),(5,5), (6,4)

n(A) = 3

$P(A)= \frac{n(A)}{n(S)}\\=\frac{3}{36}\\=\frac{1}{12}$

The probability that Rachel will win the game is: 1/12

Keywords: Probability, Sample

Learn more about probability at:

#LearnwithBrainly

7 0

3 years ago

Someone please help me with this lol… have no idea what I’m doing

Sholpan [36]

Given:

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

To find:

The quadrant of the terminal side of $\theta$ and find the value of $\sin\theta$ .

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

Here cos is positive and sine is negative. So, $\theta$ must be lies in Quadrant IV.

We know that,

$\sin^2\theta +\cos^2\theta =1$

$\sin^2\theta=1-\cos^2\theta$

$\sin \theta=\pm \sqrt{1-\cos^2\theta}$

It is only negative because $\theta$ lies in Quadrant IV. So,

$\sin \theta=-\sqrt{1-\cos^2\theta}$

After substituting $\cos \theta =\dfrac{3}{5}$ , we get

$\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}$

$\sin \theta=-\sqrt{1-\dfrac{9}{25}}$

$\sin \theta=-\sqrt{\dfrac{25-9}{25}}$

$\sin \theta=-\sqrt{\dfrac{16}{25}}$

$\sin \theta=-\dfrac{4}{5}$

Therefore, the correct option is B.

6 0

3 years ago