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Gennadij [26K]
3 years ago
6

A graphic designer wants to translate rectangle DEFG using T–1, 2(x, y). The pre-image has coordinates D(–1, 3),

Mathematics
2 answers:
Lesechka [4]3 years ago
7 0

Answer:The answer is B !

Step-by-step explanation:

Cerrena [4.2K]3 years ago
4 0

Answer:The answer is b I got it correct in my testm

Step-by-step explanation:

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<img src="https://tex.z-dn.net/?f=3q%2B2p%2B7" id="TexFormula1" title="3q+2p+7" alt="3q+2p+7" align="absmiddle" class="latex-for
kirza4 [7]

Answer:

that seem pretty hard but i think ik it

6 0
3 years ago
What is 3/4 -7/10 -3/4 and 8/10 In order from least to greatest
zaharov [31]
-3/4, -7/10, 3/4, 8/10 is least to greatest
3 0
3 years ago
Read 2 more answers
It is positive and increasing
Oduvanchick [21]
It is positive and increasing
5 0
3 years ago
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
3 years ago
47/100+3/10+47/100=/100
ira [324]

Answer:

The total is 97/100

Step-by-step explanation:

47/100+3/100=50/100 (aka 1/2)

50/100+47/100=97/100

5 0
2 years ago
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