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nikdorinn [45]
2 years ago
15

To solve an equation of the form ax=b you need to divide by the _____.

Mathematics
2 answers:
ss7ja [257]2 years ago
8 0
A. Coefficient - Apex as well
Veseljchak [2.6K]2 years ago
6 0
A is the coefficient of x in this case.

So A. is the right answer as you need to divide by the coefficient.
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David and Amanda are trying to figure out how long they can live off their $12,350 savings if they spend $240 each month. They h
Romashka [77]
David's : y = 12,350 - (240x)
Amanda's : y = 12,350 + (240x)

David's equation is correct, because their spending will be multiplied by the number of months and then subtracted from their savings


7 0
2 years ago
Read 2 more answers
Help plz my teacher won't help
lesantik [10]

Answer:

y = -3x + 9

Step-by-step explanation:

y = mx + b where m is the slope and b is the y intercept

The hypotenuse is on the same line as BC but twice as long.  So extend that line up and it will cross the y axis at (0,9).  9 will be your y intercept.

To find the slope start at C and count  up 9 and to the left 3 to get to B.

So the slope is rise/run or 9/(-3) = -3.

Final equation would be y = -3x + 9

6 0
3 years ago
Please help will give brainliest!! thank u
Goshia [24]

9514 1404 393

Answer:

  (c)  162 cm

Step-by-step explanation:

The centroid divides a median into parts with the ratio 1:2, so RL:LD = 1:2. Then RL:RD = 1:(1+2), and RD = 3RL.

  RD = 3·54 cm

  RD = 162 cm

5 0
2 years ago
Michael is constructing a circle circumscribed about a triangle. he has partially completed the construction. what should be his
earnstyle [38]

Answer:

Connect the arcs to make a perpendicular bisector

6 0
10 months 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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