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3 years ago

Need help finding the value of O. Please help..

Mathematics

1 answer:

Kruka [31]3 years ago

7 0

Answer:

O=30° or O=29°

Step By Step Explanation:

∠α =arcsin( a/c )

= arcsin(4/8)

= arcsin(0.5)

= 0.5 rad = 30° = 29°

I Hope This Helps. I Gave you 2 answers just in case 1 is Wrong. And I Hope you Understand This Calculation, I Used Different Variables and the arcsin thing is something my teacher told my class to use, and I Have no Idea what it means because my teacher never explained what it meant, Sadly.

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AnnyKZ [126]

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3 years ago

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Mashutka [201]

Answer:

6. 550

7. 70

Step-by-step explanation:

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3 0

3 years ago

Two in-line skaters, Nell and Kristin, start from the same point and skate in the same direction. Nell skates at 16 miles per ho

Paha777 [63]

First, we count how far reached the Nell:
3,5 * 16 = 56 miles

Secondly, we count how far reached the Kristin:
3,5 * 18 = 63 miles

Now we expect the difference in the distance between them:
63 - 56 = 7 miles

The results that Kristin about seven miles driven more than Nell.

8 0

3 years ago

If f=3,8 = 5, h= 10 find the value of<br>2f-g + h

Viktor [21]

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The amount that gathered interest was £22 944.00 but, before the interest started accruing, Carol had withdrawn £1000 from the account.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

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

5 0

3 years ago