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2 years ago

What relation does not have an initial value of 50

1 answer:

4 0

I guess itz B because with the rest, a number can be added to get exactly 50. Example C. y= 50x. x could be 1 which is still 50 and D. y=50-x and x could be 0 which is still 50 whiles A. y=50 remains 50

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For each of the following sequences, determine whether it converges in probability to a constant. If it does, enter the value of

nikklg [1K]

Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached images below for the step by step explanation to the question above.

6 0

2 years ago

On the first january 2014 carol invested some money in a bank account the account payes 2.5% compound interest per year on 1st j

Sladkaya [172]

Answer:

$23,360

Step-by-step explanation:

Calculation to determine how much carol originally invested in the account

First step is to divide £23517.60 by 1.025

= (23,517.60)/(1+.025)

= (23,517.60)/1.025

=$22,944

Second step is to add back the $1,000 withdrew

=$22,944+$1,000

=$23,944

Now let calculate how much carol originally invested in the account

$23,944=1.025P

Divide both side by 1.025

P=$23,944/1.025

P=$23,360

Therefore the amount that carol originally invested in the account is $23,360

4 0

2 years ago

For which pairs of functions is (f circle g) (x)?

iVinArrow [24]

Answer:

I just took the test and aint gon cap i guessed on it but the answer is B.

f(c)=2/x and g(x)=2/x

sorry I late but i thought i should still tell

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

John took all his money from his savings account. He spent $110 on a radio and 4/11 of what was left on presents for his friend

svp [43]

Answer:

$1210

Step-by-step explanation:

Let x be total amount

First John spent $110 on a radio and 4/11 of what was left on presents for his friends so he was left with

$\frac{7}{11}(x-110)=\frac{7}{11}x-70$

Then he put 2/5 of his remaining money into a checking account

$\frac{3}{5}\left(\frac{7}{11}x-70\right)$

Rest he donated to charity

$420=\frac{3}{5}\left(\frac{7}{11}x-70\right)\\\Rightarrow \left(\frac{7}{11}x-70\right)=\frac{2100}{3}\\\Rightarrow \frac{7}{11}x-70=700\\\Rightarrow \frac{7}{11}x=770\\\Rightarrow x=1210$

Hence total amount of money John originally had was $1210

8 0

2 years ago

The variable complex number z is given by z=1+cos 2θ+isin2θ,where θ takes all values in the interval −1/2π<θ<1/2π

FinnZ [79.3K]

The given complex number is
z = 1 + cos(2θ) + i sin(2θ), for -1/2π < θ < 1/2π

Part (i)
Let V = the modulus of z.
Then
V² = [1 + cos(2θ)]² + sin²(2θ)
= 1 + 2 cos(2θ) + cos²2θ + sin²2θ

Because sin²x + cos²x = 1, therefore
V² = 2(1 + cos2θ)

Because cos(2x) = 2 cos²x - 1, therefore
V² = 2(1 + 2cos²θ - 1) = 4 cos²θ
Because -1/2π < θ < 1/2π,
V = 2 cosθ PROVEN

Part ii.
1/z = 1/[1 + cos2θ + i sin 2θ]
$\frac{1}{z} = \frac{(1+cos2\theta - i\, sin2\theta)}{(1 + cos 2\theta + i\, sin 2\theta)(1+cos2\theta - i \,sin2\theta)}\\ = \frac{1+cos2\theta - i \,sin 2\theta}{(1+cos2\theta)^{2} + sin^{2}2\theta}$

The denominator is
$(1+cos2\theta)^{2}+sin^{2}2\theta \\ = 1+2cos2\theta+cos^{2}2\theta+sin^{2}2\theta \\ =2cos2\theta+2 \\ = 2(1+cos2\theta)$

Therefore
$\frac{1}{z} = \frac{1}{2} -i \frac{sin2\theta}{2(1+cos2\theta)}$

The real part of 1/ = 1/ (constant).

6 0

2 years ago