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dem82 [27]
3 years ago
9

The sum of two numbers is 78. Their difference is 32. Write a system of equations that describes this situation. Solve by elimin

ation to find the two numbers. x + y = 32 y – x = 78 50 and 24 x – y = 78 x + y = 32 54 and 24 x + y = 78 x – y = 32 55 and 23
Mathematics
1 answer:
Triss [41]3 years ago
8 0
Not sure what the extra numbers and equation are, but I'll just solve it.*Note* first number and second number isn't really assigned, just shows one of the numbers of the two*
F= first number
S= second number
F+S=78
F-S= 32
Now, you want to eliminate F, so you'll multiply the first equation by -1
-(F+S=78)
F-S=32
-F-S=-78
F-S=32
-2S=-46
S=23
Now plug in the number to one of the equations.
F+23=78
78-23= 55
The two numbers are 55 and 23.
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Find the product <br><br>(-4)(-9)(-2)​
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5 0
4 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:

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• 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
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