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2 years ago

Choose the conjecture that describes how to find the 6th term in the sequence 4, 20, 36, 52, ….

Mathematics

1 answer:

Pie2 years ago

4 0

Answer:

Step-by-step explanation:

there is a common difference between consecutive terms in the sequence, that is

20 - 4 = 36 - 20 = 52 - 36 = 16

thus to obtain the next term in the sequence add 16 to the previous term

To obtain 2 terms distant then add 16 + 16 = 32

so 6th term = 52 + 32 = 84

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What's the quotient of I? <br> 5+5i divided by 2+I

sladkih [1.3K]

$if\ "i"\ is\ the\ imaginary\ unit:\sqrt{-1}=i\Rightarrow i^2=-1\\\\then\\\\\frac{5+5i}{2+i}=\frac{5+5i}{2+i}\cdot\frac{2-i}{2-i}=\frac{10-5i+10i+5i^2}{2^2-i^2}=\frac{10+5i-5}{4+1}=\frac{5+5i}{5}=\boxed{1+i}$

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3 years ago

You deposit $300 in a savings account that pays 6% interest compounded semiannually. How much will you have at the middle of the

Otrada [13]

Answer:

Please check the explanation.

Step-by-step explanation:

a) How much will you have at the middle of the first year?

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 0.5 years

Total amount = A = ?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

substituting the values

$A=300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(0.5\right)}$

$A=300\cdot \frac{2.06}{2}$

$A=\frac{618}{2}$

$A=309$ $

Therefore, the total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 0.5 years is $ 309.00.

Part b) How much at the end of one year?

Principle P = $300

Annual rate r = 6% = 0.06 per year

Compound n = Semi-Annually = 2

Time (t in years) = 1 years

Total amount = A = ?

Using the formula

$A\:=\:P\left(1+\frac{r}{n}\right)^{nt}$

so substituting the values

$A\:=\:300\left(1+\frac{0.06}{2}\right)^{\left(2\right)\left(1\right)}$

$A=300\cdot \frac{2.06^2}{2^2}$

$A=318.27$ $

Therefore, the total amount accrued, principal plus interest, from compound interest on an original principal of $ 300.00 at a rate of 6% per year compounded 2 times per year over 1 year is $ 318.27.

7 0

3 years ago

(10,-1) slope 1/2 slope intercept form

lawyer [7]

Answer:

y=1/2x-6

Step-by-step explanation:

y-y1=m(x-x1)

y-(-1)=1/2(x-10)

y+1=1/2x-10/2

y=1/2x-5-1

y=1/2x-6

8 0

3 years ago

Tan(x+pi) + 2sin(x+pi)=0<br><br>can you help prove this and make the left side equal the right?

vladimir1956 [14]

This is not an identity but an equation. You're supposed to find the values of $x$ for which this is true. It is *not* true for all values of $x$ .

To see why:

$\tan x$ has period $\pi$ , which means $\tan(x+\pi)=\tan x$ .

The angle sum identity for $\sin x$ says that

$\sin(x+\pi)=\sin x\cos \pi+\cos x\sin\pi=-\sin x$

$\tan(x+\pi)+2\sin(x+\pi)=\tan x-2\sin x$

By definition of $\tan x$ ,

$\tan x=\dfrac{\sin x}{\cos x}$

and so

$\tan x-2\sin x=\sin x\left(\dfrac1{\cos x}-2\right)$

which is only 0 if either $\sin x=0$ (which only happens for certain values of $x$ ) or $\dfrac1{\cos x}-2=0\implies\cos x=\dfrac12$ (which also only happens for certain values of $x$ ). It's these values of $x$ you want to find.

$\sin x=0$ whenever $x$ is a multiple of $\pi$ , i.e. $x=n\pi$ for any integer $n$ .

If $0\le x$ , then $\cos x=\dfrac12$ is true for $x=\dfrac\pi3$ and $x=\dfrac{5\pi}3$ . Then to account for all other possible values, we add a multiple of $2\pi$ , so that $x=\dfrac\pi3+2n\pi$ or $x=\dfrac{5\pi}3+2n\pi$ for integers $n$ .