What is 243 written as a power of a number?

1 answer:

8 0

A to the power of b

a^b

a = base

b = exponent

SO

10^243 or 10 to the power of 243.

You can use other numbers also.

Hope i helped!!

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Write -4i+(1/4-5i)-(-3/4+8i)+17i as a complex number in the standard for

KatRina [158]

Answer:

$1+ 0i$

Step-by-step explanation:

A complex number is a number which has some real part and some imaginary part.

Standard form of a complex number is represented as

$a +bi$

Where $a$ is the real part,

and $bi$ is the imaginary part.

And $i = \sqrt{-1}$

Given complex number:

$-4i+\dfrac{1}{4}-5i)-(-\dfrac{3}{4}+8i)+17i\\\Rightarrow -4i+\dfrac{1}{4}-5i + \dfrac{3}{4}-8i+17i\\\Rightarrow \dfrac{3}{4}+\dfrac{1}{4}-4i -5i-8i+17i\\\Rightarrow \dfrac{3+1}{4}-17i+17i\\\Rightarrow \dfrac{4}{4}+0i\\\Rightarrow 1 + 0i$

Hence, the standard form is $1+ 0i$ .

8 0

2 years ago

On the 1st January 2014 Carol invested some money in a bank account.

Ghella [55]

Answer:

$\large \boxed{\text{\pounds 23 360.00}}$

Step-by-step explanation:

The formula for the accrued amount from compound interest is

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

1. Amount in account on 1 Jan 2015

(a) Data:

a = £23 517.60

r = 2.5 %

n = 1

t = 1 yr

(b) Calculations:

r = 0.025

$\begin{array}{rcl}23517.60 & = & P\left (1 + \dfrac{r}{n}\right)^{nt}\\\\& = & P\left (1 + \dfrac{0.025}{1}\right)^{1\times1}\\\\& = & P (1 + 0.025)\\ & = & 1.025 P\\P & = & \dfrac{23517.60 }{1.025} \\\\& = & 22 944.00 \\\end{array}$

The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.

She must have had £23 944 in her account on 1 Jan 2015.

(2) Amount originally invested

(a) Data

A = £23 944.00

$\begin{array}{rcl}23 944.00 & = & 1.025 P\\P & = & \dfrac{23 944.00 }{1.025} \\\\& = & \mathbf{23 360.00} \\\end{array}\\\text{Carol originally invested $\large \boxed{\textbf{\pounds23 360.00}}$ in her account.}$