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Fed [463]
3 years ago
10

Which of these is the algebraic expression for "3 less than the product of 4 and some number?"

Mathematics
1 answer:
svet-max [94.6K]3 years ago
7 0

Answer:

4s - 3

hope this helps!

Step-by-step explanation:

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What is the exact value of tan 30° ? enter your answer, as a simplified fraction, in the box?
attashe74 [19]

In this question , we have to find the value of tan 30  and we need to write the answer in simplified form .

tan x = \frac{sin x}{ cos x}

Substituting the values of sin (30) and cos(30, we will get

tan (30) = \frac{1/2}{ \sqrt{3} /2}

Multiplying both numerator and denominator by 2, we will get

tan(30 )  = \frac{1}{ \sqrt3}

TO simplify it, we need to multiply both numerator and denominator by square root 3.

tan(30) = \frac{ \sqrt 3}{ \sqrt3 * \sqrt 3} = \frac{ \sqrt 3}{3}

6 0
2 years ago
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A recipe calls for 2/3 cup of milk for 11 cookies. How many cups of milk are needed for 165 cookies​
gavmur [86]

Answer:

11 x 15 = 165, so to make 165 cookies, you would need 15 times the milk.

2/3 times 15 = 10

You would need 10 cups of milk to make 165 cookies

7 0
2 years ago
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Which of the following situations can be represented by a proportional relationship?
Zina [86]

Answer:

i think D

Step-by-step explanation:

im not great at math but yea i think D

6 0
3 years ago
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24*43x22*8x67x890x10*3x6*1x8*1
lukranit [14]

Answer:

1032x176x67x890x30x6x8

1032×176×67×890×30×6×8x^6

1.56^13×x^6

4 0
2 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
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