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gavmur [86]
3 years ago
6

If u +t=5 and u – t = 2, what is the value of (u – t)(u2 – ť)? ?

Mathematics
1 answer:
raketka [301]3 years ago
5 0

Answer:

The Heaviside step function, or the unit step function, usually denoted by H or θ (but sometimes u, 1 or ), is a discontinuous function, named after Oliver Heaviside (1850–1925), whose value is zero for negative arguments and one for positive arguments.

Step-by-step explanation:

L at time ť will meet at the point z " given by the equations : z " = - ( c + U ) t ... + B ) t = ť + t " = 2y - t = tı ( 93 ) Therefore , we do not recover the value 2t = 2L / c ... if the extinction time is not zero in equation 92 , one can verify the equality t ' + t ... the mirror located on the arm covers the distance Ut ' during the time t

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y+2=-(x+8)
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Why is 2(x+9) + 4x an expression and not an equation?​
s2008m [1.1K]

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because it doesn't have an equal sign.

Step-by-step explanation:

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It cost Naomi $7.70 to send 154 text messages. How many text messages did she send
s2008m [1.1K]

You have to find the unit rate

SO..

7.70 / 154 = 0.05

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So then you divide $ 5.25 by the unit rate to get the number of text

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To check it you just multiply the unit rate by the number of text you got and you show come up with $ 5.25

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4 0
3 years ago
Show that 3 · 4^n + 51 is divisible by 3 and 9 for all positive integers n.
-BARSIC- [3]

Answer:

Step-by-step explanation:

Hello, please consider the following.

3\cdot 4^n+51=3\cdot 4^n+3\cdot 17=3(4^n+17)

So this is divisible by 3.

Now, to prove that this is divisible by 9 = 3*3 we need to prove that

4^n+17 is divisible by 3. We will prove it by induction.

Step 1 - for n = 1

4+17=21= 3*7 this is true

Step 2 - we assume this is true for k so 4^k+17 is divisible by 3

and we check what happens for k+1

4^{k+1}+17=4\cdot 4^k+17=3\cdot 4^k + 4^k+17

3\cdot 4^k is divisible by 3 and

4^k+17 is divisible by 3, by induction hypothesis

So, the sum is divisible by 3.

Step 3 - Conclusion

We just prove that 4^n+17 is divisible by 3 for all positive integers n.

Thanks

4 0
3 years ago
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