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Vinvika [58]
3 years ago
6

Find the sum or difference

Mathematics
2 answers:
enot [183]3 years ago
5 0
To add matrices, add the terms in the corresponding boxes. The resulting matrix will be your answer:

\left[\begin{array}{ccc}-8&-1&7\\0&9&2\end{array}\right] +   \left[\begin{array}{ccc}-2&0&-2\\-4&5&-1\end{array}\right] =   \left[\begin{array}{ccc}-10&-1&5\\-4&14&1\end{array}\right]
Solnce55 [7]3 years ago
4 0
Hi there!

For adding th' matrices.
Add following terms in corresponding boxes.

\left[\begin{array}{ccc}-8&-1&7\\0&9&2\end{array}\right] + \left[\begin{array}{ccc}-2&0&-2\\-4&5&-1\end{array}\right] = \left[\begin{array}{ccc}-10&-1&5\\-4&14&1\end{array}\right]

~ Hope it helps!
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He makes $316 a week and $23 for every sale he makes. He has to pay his rent this week that is $500, so he will need to make at
Lyrx [107]

let x be the number of sales Hayward needs to make

316+23x > or equals to 500

23x> or equals to 184

x> or equals to 8

4 0
3 years ago
Anyone know this?????
kifflom [539]

Answer:

WZX=57

Step-by-step explanation:

90-33 = 57

You don't have to use the equation unless you want to check yourself.

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3 years ago
The bar graph below shows the attendance at school dances. The ticket per price for each dance is $5 per student. Which two danc
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It's B because all the students that attended added together is 320 and 37.5 of 320 is 120

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3 years ago
Does anybody know how to do time with exponential decay
My name is Ann [436]
<span>From the message you sent me:

when you breathe normally, about 12 % of the air of your lungs is replaced with each breath. how much of the original 500 ml remains after 50 breaths

If you think of number of breaths that you take as a time measurement, you can model the amount of air from the first breath you take left in your lungs with the recursive function

b_n=0.12\times b_{n-1}

Why does this work? Initially, you start with 500 mL of air that you breathe in, so b_1=500\text{ mL}. After the second breath, you have 12% of the original air left in your lungs, or b_2=0.12\timesb_1=0.12\times500=60\text{ mL}. After the third breath, you have b_3=0.12\timesb_2=0.12\times60=7.2\text{ mL}, and so on.

You can find the amount of original air left in your lungs after n breaths by solving for b_n explicitly. This isn't too hard:

b_n=0.12b_{n-1}=0.12(0.12b_{n-2})=0.12^2b_{n-2}=0.12(0.12b_{n-3})=0.12^3b_{n-3}=\cdots

and so on. The pattern is such that you arrive at

b_n=0.12^{n-1}b_1

and so the amount of air remaining after 50 breaths is

b_{50}=0.12^{50-1}b_1=0.12^{49}\times500\approx3.7918\times10^{-43}

which is a very small number close to zero.</span>
5 0
3 years ago
Mary has 10/12 of ube cake she shared 7/12 to ana. what part of cake was left to mary?​
kow [346]

Answer:

3/12

Step-by-step explanation:

10-7 is 3, so its 3/12.

5 0
3 years ago
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