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Ivanshal [37]
3 years ago
14

Please help with this page

Mathematics
2 answers:
REY [17]3 years ago
7 0
1. )2.7
2.)8.875
3.)9.
4.)4.
Ulleksa [173]3 years ago
7 0
There are 14 quarters in $3.50
1||97.2/3.6=2.7
2||6.39/0.72=8.875
3||0.81/0.09= 9.0
4||1.08/0.27=4.0
5||Nathan is correct since they put the decimal point one space ahead of where it was supposed to be and that the decimal should be in between the 1 and 9 by estimation.
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Find the area of the shape.
vesna_86 [32]

Answer:

what was the answer?? I am also needing to know!

8 0
2 years ago
A circular fish pond is shown below. What<br> is the circumference of the pond? The radius is 18.5cm
alisha [4.7K]

Answer:

about 116.18

Step-by-step explanation:

Circumference is 2PI*radius

So it should equal 37*3.14 which equals about 116.18

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The temperature in the desert can
Tanya [424]

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There is an 144 degree difference.

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112=(-32)+x

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4 0
3 years ago
If g(x) = x - 3 and f(x) = 3 - x2; find f(g(-2))=
const2013 [10]

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Step-by-step explanation:

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5 0
3 years ago
Is matrix multiplication comutative?
konstantin123 [22]

Answer:

Matrix multiplication is not conmutative

Step-by-step explanation:

The matrix multiplication can be performed if the number of columns of the first matrix is equal to the number of rows of the second matrix

Let A with dimension mxn and B with dimension nxp represent two matrix

The multiplication of A by B is a matrix C with dimension mxp, but the multiplication of B by A is can't be calculated because the number of columns of B is not the number of rows of A. Therefore, you can notice that is not conmutative in general.

But even if the multiplication of AB and BA is defined (For example if A and B are squared matrix of 2x2) the multiplication is not necessary conmutative.

The matrix multiplication result is a matrix which entries are given by dot product of the corresponding row of the first matrix and the corresponding column of the second matrix:

A=\left[\begin{array}{ccc}a11&a12\\a21&a22\end{array}\right]\\B= \left[\begin{array}{ccc}b11&b12\\b21&b22\end{array}\right]\\AB = \left[\begin{array}{ccc}a11b11+a12b21&a11b12+a12b22\\a21b11+a22b21&a21b12+a22b22\end{array}\right]\\\\BA=\left[\begin{array}{ccc}b11a11+b12a21&b11a12+b12a22\\b21a11+b22ba21&b21a12+b22a22\end{array}\right]

Notice that in general, the result is not the same. It could be the same for very specific values of the elements of each matrix.

6 0
3 years ago
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