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Lana71 [14]
3 years ago
7

Pleas please help! Solve for d. 3/15 - d/5 = 5/6 + d/3

Mathematics
2 answers:
ololo11 [35]3 years ago
7 0
\frac{3}{15}-<u />\frac{d}{5}-(\frac{5}{6}+\frac{d}{3}=0

Step 1:<u />\frac{5}{6}+\frac{d}{3}=<u /><u />\frac{3}{15}- \frac{d}{5}- \frac{(2d+5)}{6}=0

Step 2:\frac{1}{5}- \frac{d}{5}= \frac{(1-d)}{5}- \frac{(2d+5)}{6}=0

Step 3:\frac{(1-d)*6-(2d+5)*5}{30}= \frac{-1*(16d+19)}{30}

Step 4:16d= -19:Divide \ both \ sides \ by \ 16:d= \frac{-19}{16}= -1.188

Your answer is <u />d\boxed{\boxed={ \frac{-19}{16}= -1.1888}}}}
Papessa [141]3 years ago
3 0
3/15 -(d/5)=5/6+(d/3)
-(d/5)-(d/3)=5/6-3/15

-(8d/15)=19/30
-8d=19/2
-d=76
d=-76
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A= [2, 3; 2, 1], B=[3; 5]. Find AB &amp; BA if possible
oksano4ka [1.4K]

Answer:

The value of AB is \left[\begin{array}{ccc}21\\11\end{array}\right] and it's not possible to multiply BA.

Step-by-step explanation:

Consider the provided matrices.

A=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right], B=\left[\begin{array}{ccc}3\\5\end{array}\right]

Two matrices can be multiplied if and only if first matrix has an order m × n and second matrix has an order n × v.

Multiply AB

Matrix A has order 2 × 2  and matrix B has order 2 × 1. So according to rule we can multiply both the matrix as shown:

AB=\left[\begin{array}{ccc}2&3\\2&1\end{array}\right] \left[\begin{array}{ccc}3\\5\end{array}\right]

AB=\left[\begin{array}{ccc}2\times 3+3\times 5\\2\times 3+1\times 5\end{array}\right]

AB=\left[\begin{array}{ccc}6+15\\6+5\end{array}\right]

AB=\left[\begin{array}{ccc}21\\11\end{array}\right]

Hence, the value of AB is \left[\begin{array}{ccc}21\\11\end{array}\right]

Now calculate the value of BA as shown:

Multiply BA

Matrix B has order 2 × 1  and matrix A has order 2 × 2. So according to rule we cannot multiply both the matrix.

We can multiply two matrix if first matrix has an order m × n and second matrix has an order n × v.

That means number of column of first matrix should be equal to the number of rows of second matrix.

Hence, it's not possible to multiply BA.

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