The cost of 5 cans of dog food is 4.35. At this price ,how much do 11 cans of dog food cost

Mathematics

1 answer:

Setler [38]3 years ago

5 0

Answer:

$9.57

Step-by-step explanation:

You have to do 4.35 divided by 5 to get how much 1 can cost. You will find out that 1 can cost $0.87. Next you will take how much one cost and mutiply it by 11. When you do that you will get 9.57. That is how you will get $9.57 as your answer.

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I need to know the answer please help! and thank you!

Snowcat [4.5K]

Answer:

v/2

Step-by-step explanation:

First, if we expand the expression we know that it is equal to (4/8)*(v^3/v^2)

We can simplify the numbers to get (1/2)*(v^3/v^2)

We know that division of exponents is just subtraction of them, so v³/v² = v^(3-2) = v^1 = v

So, the answer is 1/2 * v or v/2

I hope this helps!

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3 years ago

What is the probability of getting at least one 6 in four rolls of a single fair die?

Alik [6]

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3 years ago

Read 2 more answers

A = ???? 4 −2

irinina [24]

Answer:

1. The matrix A isn't the inverse of matrix B.

2. |B|=12, |A|=12

Step-by-step explanation:

1. We want to know if matrix A is the inverse of matrix B, this means that if you do the product between B and A you have to obtain the identity matrix.

We have:

$A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]$

and

$B=\left[\begin{array}{cc}3&2\\1&4\end{array}\right]$

A and B are 2×2 matrices (2 rows and 2 columns), if you multiply them you have to obtain a 2×2 matrix.

Then if A is the inverse of B:

$B.A=I$

Where,

$I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$

Observation:

If you have two matrices:

$A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A.B=\left[\begin{array}{cc}(a.e+b.g)&(a.f+b.h)\\(c.e+d.g)&(c.f+d.h)\end{array}\right]$

Now:

$B.A=\left[\begin{array}{cc}3&2\\1&4\end{array}\right].\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}4.3+(-2).1&4.2+(-2).4\\(-1).3+3.1&(-1).2+3.4\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}12-2&8-8\\-3+3&-2+12\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]$