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Debora [2.8K]
3 years ago
10

(7) (10) = (x) (4) I need help with this please anyone?

Mathematics
1 answer:
otez555 [7]3 years ago
5 0
(7)(10)=(x)(4)
70=(x)(4)
70/4 = (x)(4) / 4
17.5=x
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I use 1.48 inches of ribbon to make 4 necklaces. How much ribbon did I use for each necklace?
grin007 [14]
0.37 inches on each necklace or 37/100.

You just need to do 1.48 divided by 4.

Hope this helps !!!
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7 0
3 years ago
Kurt used the rule add 4, subtract 1 to generate a pattern. The first term in his pattern is 5. Write a number that could be in
Rainbow [258]
5, 8, 11, 14, 17, 20, 23
4 0
2 years ago
In matrix multiplication, such as AB=C, the columns of B form the rows of C. why is this?
Gnesinka [82]
Let's work with 2-by-2 matrices so we're on the same page. The ideas will work for any appropriate matrices.

From the rule of matrix multiplication, we see:
\left[\begin{array}{cc}a_{11} & a_{12} \\a_{21} & a_{22} \end{array}\right] \left[\begin{array}{cc}b_{11} & b_{12} \\b_{21} & b_{22} \end{array}\right] = \left[\begin{array}{cc} a_{11}b_{11} + a_{12}b_{21} & a_{11}b_{12} + a_{12}b_{22} \\ a_{21}b_{11} + a_{22}b_{21} & a_{21}b_{12} + a_{22} b_{22} \end{array}\right]
As you noted, we see the columns of B contributing to the rows of C. The question is, why would we ever have defined matrix multiplication this way?

Here's a nontraditional way of feeling this connection. We can define matrix multiplication as "adding multiplication tables." A multiplication table is made by starting with a column and a row. For example,
\begin{array}{ccc} {} & 1 & 2 \\ 1 & {} & {} \\ 2 & {} & {} \end{array}
We then fill this table in by multiplying the row and column entries:
\begin{array}{ccc} {} & [1] & [2] \\ 1| &1 & 2 \\ 2| & 2 &4 \end{array}
It's then reasonable to say that given two matrices A and B, we can construct multiplication tables by taking the columns of A and pairing them with the rows of B:
\left[\begin{array}{cc}a_{11} & a_{12} \\a_{21} & a_{22} \end{array}\right] \left[\begin{array}{cc}b_{11} & b_{12} \\b_{21} & b_{22} \end{array}\right]

= \begin{array}{cc} {} & \left[\begin{array}{cc} b_{11} & b_{12}\end{array} \right]\\ \left[\begin{array}{c} a_{11} \\ a_{21} \end{array} \right] \end{array} +\begin{array}{cc} {} & \left[\begin{array}{cc} b_{21} & b_{22}\end{array} \right]\\ \left[\begin{array}{c} a_{12} \\ a_{22} \end{array} \right] \end{array}

= \left[\begin{array}{cc} a_{11} b_{11} & a_{11} b_{12} \\ a_{21} b_{11} & a_{21} b_{12} \end{array} \right] + \left[\begin{array}{cc} a_{12} b_{21} & a_{12} b_{22} \\ a_{22} b_{21} & a_{22} b_{22} \end{array} \right]

Adding these matrices together, we get the exact same expression as the traditional definition. 




5 0
3 years ago
F(x) = – 2x<br> g(x) = 8x2- 5x+7<br> Find (f.g)(x).
mr_godi [17]

f(x) = -2x

-2x = 0

x = 0/-2

x = 0

Now putting 0 as x in g(x)

g(x) = 8x^2 - 5x + 7

g(0) = 8(0)^2-5(0)+7

= 0 - 0 + 7

= 7

Answered by Gauthmath must click thanks and mark brainliest

4 0
2 years ago
Help don’t understand
Misha Larkins [42]

(4*<em>x</em>)-6 because 6 less than a product of 4 and <em>x</em> would mean you have to have a product of 4 and x before you subtract 6

8 0
3 years ago
Read 2 more answers
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