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3 years ago

Can someone please answer this for me

Mathematics

2 answers:

GREYUIT [131]3 years ago

6 0

Answer: 15

Step-by-step explanation:

Anton [14]3 years ago

6 0

Answer:

A = 30m + 35

Step-by-step explanation:

let A be the amount saved and m represent month

Note the difference in the amount saved is constant, that is

185 - 155 = 155 - 125 = 125 - 95 = 95 - 65 = 30

We can express this as

A = 30m + c

To find c use any pair from the table and substitute into the equation

using (3, 125) then

125 = (3 × 30) + c

125 = 90 + c ( subtract 90 from both sides )

c = 125 - 90 = 35, thus

A = 30m + 35 ← linear relationship

As a check

When m = 4, A should be 155

A = (4 × 30) + 35 = 120 + 35 = 155 ← correct

Linear function representing the table is

A(m) = 30m + 35

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3 years ago

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Step-by-step explanation:

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3 years ago

M1 Matrix is:

harkovskaia [24]

Answer:

(a)First Row, First Column =1

(b)First Row, second Column =0

(c)Second Row, First Column =0

(d)Second Row, second Column =1

Step-by-step explanation:

Given matrix $M=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)$

The Inverse of a 2X2 matrix

$A=\left(\begin{array}{ccc}a&b\\c&d\end{array}\right)$

can be found using the following:

$A^{-1}=\dfrac{1}{ad-bc} \left(\begin{array}{ccc}d&-b\\-c&a\end{array}\right)$

Therefore:

$M^{-1}=\dfrac{1}{(5*-5)-(3*-8)} \left(\begin{array}{ccc}5&-3\\8&-5\end{array}\right)\\=-1\left(\begin{array}{ccc}5&-3\\8&-5\end{array}\right)\\=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)$

Next, we find the product $M^{-1}M$

$M^{-1}M=\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)\left(\begin{array}{ccc}-5&3\\-8&5\end{array}\right)\\=\left(\begin{array}{ccc}-5*-5+3*-8&-5*3+3*5\\-8*-5+5*-8&-8*3+5*5\end{array}\right)\\=\left(\begin{array}{ccc}1&0\\0&1\end{array}\right)$

Therefore:

(a)First Row, First Column =1

(b)First Row, second Column =0

(c)Second Row, First Column =0

(d)Second Row, second Column =1

NOTE: The multiplication of a matrix and its inverse always gives the identity matrix as seen above,

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