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3 years ago

Help me YALL lol!!!!!!!!!!!!!!!

Mathematics

1 answer:

slavikrds [6]3 years ago

3 0

Answer:

15=2x

There are fifteen 1 blocks and two x blocks, which makes 15=2x(or 2x=15)

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Can somebody help me please

SpyIntel [72]

<h2>Answer:</h2><h3>f(4) = 122</h3>

________________________________________________

<h3>Calculate</h3>

$Substitute\ x=4\ into\\ f(x)=7x^2+4x-6$

_________________________________________________

<h3>Substitute</h3>

$f(4)=7\times 4^2+4\times 4-6$

___________________________________________________

<h3>Calculate the power</h3>

$f(4)=7\times 16+4\times 4-6$

___________________________________________________

<h3>Calculate the product or quotient</h3>

$f(4)=112+4\times 4-6\\ f(4)=112+16-6$

___________________________________________________

<h3>Calculate the sum or difference</h3>

$f(4)=128-6\\ f(4)=122$

<em>I hope this helps you</em>

6 0

3 years ago

Find the sum of 2 $ + (-1) + 39

kozerog [31]

Answer:

C.5

Step-by-step explanation:

4 0

3 years ago

The perimeter of a rectangle is 56 inches. The length is 4 more than 3 times the width. What are the dimensions of the rectangle

VashaNatasha [74]

Let the length be x and the width be y, then
2x + 2y = 56
x + y = 28 . . . (1)
x = 4 + 3y . . . (2)

Putting (2) into (1) gives
4 + 3y + y = 28
4y = 28 - 4 = 24
y = 24/4 = 6
x = 4 + 3(6) = 4 + 18 = 22

Therefore, length is 22 inches and width is 6 inches.

5 0

4 years ago

Evaluate 2xy when x equals -1/4 and y equals 3

umka21 [38]

The answer is -1.5
2(-1/4•3) is what it looks like after everything is put in. Hope this helps

5 0

3 years ago

Read 2 more answers

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

4 years ago