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Rus_ich [418]
3 years ago
15

Can you please help me?

Mathematics
2 answers:
ser-zykov [4K]3 years ago
8 0

Answer:

idek

Step-by-step explanation:

srry

Yuliya22 [10]3 years ago
8 0

Step-by-step explanation:

g(2)= -11

g(4)= -9

g(6)= -7

g(8)= -5

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What would be the answer too 900ft2+ 144ft2???
In-s [12.5K]

Answer: 1044ft

Step-by-step explanation:

900ft2 + 144ft2

= 1044ft2 (square root)

1044ft

hope this helps :))

5 0
3 years ago
The calculation of property tax is based on the
Tresset [83]
The assessment rate is a uniform percentage and varies by tax jurisdiction, and could be any percentage below 100%. After getting the assessed value, it is multiplied by the mill levy to determine your taxes due. For example, suppose the assessor determines your property value is $500,000 and the assessment rate is 8%.
5 0
3 years ago
Evaluate the following integrals: 1. Z x 4 ln x dx 2. Z arcsin y dy 3. Z e −θ cos(3θ) dθ 4. Z 1 0 x 3 √ 4 + x 2 dx 5. Z π/8 0 co
Zigmanuir [339]

Answer:

The integrals was calculated.

Step-by-step explanation:

We calculate integrals, and we get:

1) ∫ x^4 ln(x) dx=\frac{x^5 · ln(x)}{5} - \frac{x^5}{25}

2) ∫ arcsin(y) dy= y arcsin(y)+\sqrt{1-y²}

3) ∫ e^{-θ} cos(3θ) dθ = \frac{e^{-θ} ( 3sin(3θ)-cos(3θ) )}{10}

4) \int\limits^1_0 {x^3 · \sqrt{4+x^2} } \, dx = \frac{x²(x²+4)^{3/2}}{5} - \frac{8(x²+4)^{3/2}}{15} = \frac{64}{15} - \frac{5^{3/2}}{3}

5)  \int\limits^{π/8}_0 {cos^4 (2x) } \, dx =\frac{sin(8x} + 8sin(4x)+24x}{6}=

=\frac{3π+8}{64}

6)  ∫ sin^3 (x) dx = \frac{cos^3 (x)}{3} - cos x

7) ∫ sec^4 (x) tan^3 (x) dx = \frac{tan^6(x)}{6} + \frac{tan^4(x)}{4}

8)  ∫ tan^5 (x) sec(x)  dx = \frac{sec^5 (x)}{5} -\frac{2sec^3 (x)}{3}+ sec x

6 0
3 years ago
What’s 5x divided by 6=20
lilavasa [31]

Answer:

24

Step-by-step explanation:

5x/6=20

Multiple 20 by 6 then you get 120.

Then divide 120 by 5 which is 24.

5 0
3 years ago
Read 2 more answers
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
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