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3 years ago

In triangles △ABC and △DEF,

Mathematics

1 answer:

wel3 years ago

7 0

Answer:

BC = 4.8
ED = 1.1
DF = 1.6

Step-by-step explanation:

Since angles A and E correspond, as well as angles C and F, we can say ...

ΔABC ~ ΔEDF

Then the ratio of side lengths of ΔABC to those of ΔEDF is ...

AC/EF = 6/2 = 3

That means ...

ED/AB = 1/3

ED = AB·(1/3) = 3.3·(1/3) = 1.1

For the remaining sides, we have the relation

3·DF = BC

3·(BC -3.2) = BC

2BC - 9.6 = 0 . . . eliminate parentheses, subtract length BC

BC -4.8 = 0 . . . . . divide by 2

BC = 4.8 . . . . . . . . add 4.8

DF = BC·(1/3) = 1.6

The unknown side lengths are BC = 4.8, DE = 1.1, DF = 1.6.

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The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.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

3. Summary

1 Jan 2014 P = £23 360.00

1 Jan 2015 A = 23 944.00

Withdrawal = <u> -1 000.00 </u>

P = 22 944.00

1 Jan 2016 A = £23 517.60

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