$if \: log(x ) = 3 \\ log_{10}(x) = 3 \\ {10}^{3} = x \\ therfore \to \: \boxed{x = {10}^{3} } \\ \\ if \: log(y ) = 2 \\ log_{10}(y) = 2 \\ {10}^{2} = y \\ therfore \to \: \boxed{y= {10}^{2} }$

Step-by-step explanation:

$x^2 -7xy =3000y \: then \to \\ x^2 =3000y + 7xy \\ {(10 ^{3} )}^{2} = 3000(10)^{2} + 7( {10}^{3} \times {10}^{2}) \\ {10}^{6} = 3000(10)^{2} + 7( {10}^{3} \times {10}^{2}) \\ {10}^{6} = 3000(10)^{2} + 7( {10}^{5} ) \\ {10}^{6} = 300000 + 700000 \\ \boxed{{10}^{6} = 1000000} \\ hence \: the \: equation \to \: x^2 -7xy =3000y \\ \boxed{ is \: corret}$

♨Rage♨

3 0

3 years ago

A broker charges customers a fixed fee of £150 plus 5.50% of the amount of all investments made. If a customer invests £40 000 t

Romashka [77]

Answer:

Total payable brokerage = £2350

Step-by-step explanation:

Given that:

Fixed fee charged by broker = £150

Percent of fee charged = 5.50%

Amount invested by customer = £40,000

Fee charged = $\frac{5.50}{100}*40000$

Fee charged = 5.50 * 400 = £2200

Total brokerage payable = £150 + £2200 = £2350

Hence,

Total payable brokerage = £2350

8 0

3 years ago