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blondinia [14]
3 years ago
15

James age is 3 years less than twice Tori's age the sum of their ages is 30 find their ages

Mathematics
1 answer:
blagie [28]3 years ago
7 0
Error// Error// Error// Error// Error// Error//
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In the theory of learning, the rate at which a subject is memorized is assumed to be proportional to the amount that is left to
babymother [125]

Answer:

dA/dt = k1(M-A) - k2(A)

Step-by-step explanation:

If M denote the total amount of the subject and A is the amount memorized, the amount that is left to be memorized is (M-A)

Then, we can write the sentence "the rate at which a subject is memorized is assumed to be proportional to the amount that is left to be memorized" as:

Rate Memorized = k1(M-A)

Where k1 is the constant of proportionality for the rate at which material is memorized.

At the same way, we can write the sentence: "the rate at which material is forgotten is proportional to the amount memorized" as:

Rate forgotten = k2(A)

Where k2 is the constant of proportionality for the rate at which material is forgotten.

Finally, the differential equation for the amount A(t) is equal to:

dA/dt = Rate Memorized - Rate Forgotten

dA/dt = k1(M-A)  - k2(A)

7 0
3 years ago
The first person to answer correctly will get brainliest 5 stars AND a thank you!!!!
Travka [436]

Answer:

A translation

Step-by-step explanation:

:)

5 0
3 years ago
Whajsoska help pls omgggggg ​
noname [10]

Answer:

I can answer the first one the other 2 I can't make out.

Step-by-step explanation:

Firstly,

1) They are the same since they add one to the row every time

2) They are different because Megan's pattern is that they only count the first column in orange, and Kyle's pattern counts the number of blocks in each row.

3) They have the same number of blocks just in different colors in rows and columns.

8 0
2 years ago
Given $m\geq 2$, denote by $b^{-1}$ the inverse of $b\pmod{m}$. That is, $b^{-1}$ is the residue for which $bb^{-1}\equiv 1\pmod
goblinko [34]

(2+3)^{-1}\equiv5^{-1}\pmod7 is the number <em>L</em> such that

5L\equiv1\pmod7

Consider the first 7 multiples of 5:

5, 10, 15, 20, 25, 30, 35

Taken mod 7, these are equivalent to

5, 3, 1, 6, 4, 2, 0

This tells us that 3 is the inverse of 5 mod 7, so <em>L</em> = 3.

Similarly, compute the inverses modulo 7 of 2 and 3:

2a\equiv1\pmod7\implies a\equiv4\pmod7

since 2*4 = 8, whose residue is 1 mod 7;

3b\equiv1\pmod7\implies b\equiv5\pmod7

which we got for free by finding the inverse of 5 earlier. So

2^{-1}+3^{-1}\equiv4+5\equiv9\equiv2\pmod7

and so <em>R</em> = 2.

Then <em>L</em> - <em>R</em> = 1.

6 0
3 years ago
(7x^2-2) - (2x^2 - 5x + 3)
ehidna [41]

(7x^2-2) - (2x^2 - 5x + 3)

Answer:

5(x^2+x-1)

4 0
3 years ago
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