What is the final amount if 943 is decreased by 13% followed by a further 6% decrease?

Mathematics

1 answer:

3241004551 [841]3 years ago

7 0

Answer:

The final amount after decreasing upto 2 DP is 771.18

Step-by-step explanation:

Amount given is 943

13% of ammount will be = 943 x 13%

= $943\times (\frac{13}{100} )\\943 \times 0.13\\122.59$

Decreasing 13% from 943 we get = 943-122.59

=820.41

Now the ammount = 820.41

for finding final amount

find

6% of amount (820.41) we get = 820.41x 6%

= $820.41\times( \frac{6}{100} )\\820.41\times 0.06\\49.23$

Decreasing 6% from 820.41 we get final amount= 820.41- 49.23

=771.18

so the final amount is 771.18

Keywords: Percentage

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Prove that if $w,z$ are complex numbers such that $|w|=|z|=1$ and $wz\ne -1$, then $\frac{w+z}{1+wz}$ is a real number.

mr_godi [17]

Answer:

See proof below

Step-by-step explanation:

Let $r=\frac{w+z}{1+wz}$ . If w=-z, then r=0 and r is real. Suppose that w≠-z, that is, r≠0.

Remember this useful identity: if x is a complex number then $x\bar{x}=|x|^2$ where $\bar{x}$ is the conjugate of x.

Now, using the properties of the conjugate (the conjugate of the sum(product) of two numbers is the sum(product) of the conjugates):

$\frac{r}{\bar{r}}=\frac{w+z}{1+wz} \left(\frac{1+\bar{w}\bar{z}}{\bar{w}+\bar{z}}{\right)$

= $\frac{(w+z)(1+\bar{w}\bar{z})}{(1+wz)(\bar{w}+\bar{z})}=\frac{w+z+w\bar{w}\bar{z}+z\bar{z}\bar{w}}{\bar{w}+\bar{z}+\bar{w}wz+\bar{z}zw}=\frac{w+z+w+|w|^2\bar{z}+|z|^2\bar{w}}{\bar{w}+\bar{z}+|w|^2z+|z|^2w}=\frac{w+z+\bar{z}+\bar{w}}{\bar{w}+\bar{z}+z+w}=1$