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Sav [38]
3 years ago
15

Can someone please clearly explain how to correctly work out Q3? Much appreciated

Mathematics
1 answer:
vfiekz [6]3 years ago
8 0

the equation for this is:

value = 36000*(0.92)^t

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Solve each of the following inequalities for x. a. 3+x>8
Leni [432]

Answer:

x > 5

Step-by-step explanation:

3+x>8

Subtract 3 from each side

3-3+x>8-3

x > 5

6 0
3 years ago
Helpp me pleasee correct answers! (10 points )
Softa [21]

Answer:  Approximately 30.4 degrees

=========================================================

Work Shown:

324 - 1.6 = 322.4 is the height of the right triangle. This is the opposite side compared to the angle theta.

The adjacent side is 550 since it is the leg closest to the angle theta.

We'll use the tangent rule to connect the two sides

tan(angle) = opposite/adjacent

tan(theta) = 322.4/550

theta = arctan(322.4/550)

theta = 30.3780566013696

theta = 30.4

Your teacher didn't provide any rounding instructions, but I rounded to one decimal place since the value 1.6 is to one decimal place.

5 0
3 years ago
-8 over -2 by the power of three Minus 3 times -2 times 4
Crazy boy [7]

Answer:88


Step-by-step explanation:

You can simplify -8/-2 its 4 then take 4 to the 3rd power and you'll get 64 the take 3×-2 and you'll get -6 then multiply -6×4 and you'll get -24 then subtract 64 - (-24) which would end up being plus a positive and you'll get 88

4 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
2 years ago
Choose either Multiplication and Division
Sliva [168]

Answer:

multiplication and divisioni gusss cause area will be give amd all" × " u have to put in "÷" fiorm

6 0
2 years ago
Read 2 more answers
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