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melomori [17]
3 years ago
14

Someone pls help this is an emergency

Mathematics
2 answers:
harina [27]3 years ago
7 0

Answer:

3,2 and -2

because you can't make denominator zero

and only these will make deno. zero

viva [34]3 years ago
5 0
3,2,-2 you set the bottom of the fraction equal to 0 then solve
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FIND THE DIFFERENCE (5x2 + 2x + 11) − (7 + 4x − 2x2)
never [62]
The answer is D. 7x squared - 2x + 4 
6 0
2 years ago
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Tom and David get their haircut at the barber shop on Saturdays.Tom gets his hair cut every 3 weeks. David gets his hair cut eve
MrMuchimi

Answer:

On Saturday, when 12 weeks passed

Step-by-step explanation:

Tom gets his hair cut every 3 weeks.

David gets his hair cut every 4 weeks.

Find the least common multiple of numbers 3 and 4:

3=3\\ \\4=2\cdot 2\\ \\LCM(3,4)=3\cdot 2\cdot 2=12

This means that Tom and David will get their hair cut on the same day at 12th week (if the first week when they started has number 0).

You can think in another way:

Tom gets his hair cut every 3 weeks. He got his hair cut after 3 weeks, after 6 weeks, after 9 weeks, after 12 weeks passed.

David gets his hair cut every 4 weeks. He got his hair cut after 4 weeks, after 8 weeks, after 12 weeks passed.

6 0
3 years ago
Help needed asap idk anything
Licemer1 [7]

Answer:

49x +196x.... or 245x

Step-by-step explanation:

8 0
3 years ago
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If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.
Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have S_1=3\times2^{1-1}+2(-1)^1=3-2=1, which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}

and

S_k=3\times2^{k-1}+2(-1)^k

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}

From the given recurrence, we know

S_{k+1}=S_k+2S_{k-1}

so that

S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)

S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}

S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)

S_{k+1}=3\times2^k-2(-1)^k

S_{k+1}=3\times2^k+2(-1)(-1)^k

\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}

which is what we needed. QED

6 0
2 years ago
A chair marked with 1800 was sold to a customur for 1504 .Find the rate of discount allowed on the chair
mihalych1998 [28]

Answer:

16.44%

Step-by-step explanation:

3 0
2 years ago
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