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3 years ago

Mid-Chapter Check

Mathematics

1 answer:

Nostrana [21]3 years ago

4 0

An equation: 2x=3
an expression: 2x+5

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Am I correct? I added them all.

max2010maxim [7]

Yes you are correct.

6 0

3 years ago

Read 2 more answers

Use the graph below for this question: graph of parabola going through negative 5, 1 and negative 4, negative 4. What is the ave

Gekata [30.6K]

Use the slope formula:
(y2 - y1)/(x2 - x1)
(x1, y1) = (-5, 1)
(x2, y2) = (-4, -4)

6 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 S₁ = 1.

• Assume the closed form is correct for all n up to n = k. 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 n = k + 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

Please help idk nothing

Rainbow [258]

Convert 9 1/2 into a mixed fraction: 19/2.
Now, multiply with 6 which in fraction form is 6/1
19/2 times 6/1 equals 114/2

Now you can divide 114/2 which is 57 inches squared.

Have a good day ! :)

7 0

3 years ago

Which expression is equivalent to 3 (4x - 5)?

Rufina [12.5K]

The answer is D 12x-15

6 0

3 years ago