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photoshop1234 [79]

2 years ago

15

Kens home supply charges

1 answer:

OleMash [197]2 years ago

8 0

IDK......Don't have a full question!

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Polynomial division help pls!!

lara31 [8.8K]

Just add the numerators and the answer becomes (4/x–4)

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3 years ago

28.5 g of iron shot is added to a graduated cylinder containing 45.50 mL of water. The water level rises to the 49.10 mL mark, f

Andrej [43]

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7.92

Step-by-step explanation:

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2 years ago

What is the correct solution set for the following graph?

Serhud [2]

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Step-by-step explanation:

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3 years ago

Can SOMEBODYYY please do the maths!!! PLeassssssssssssseeeeeeee

Anna11 [10]

Answer:

<h2>see below</h2>

Step-by-step explanation:

<h3>Question-6:</h3>

we are given a equation

$\sf \displaystyle \: \log_{4}( - x) + \log_{4}( 6 - x) = 2$

to solve so

recall logarithm multiplication law:

$\sf \displaystyle \: \log_{4}( - x \times (6 - x)) = 2$

simplify multiplication:

$\sf \displaystyle \: \log_{4}( - 6 x + {x}^{2} ) = 2$

remember $\displaystyle \log_{4}(4^2)=2$

so

$\sf \displaystyle \: \log_{4}( - 6 x + {x}^{2} ) = \log_{4}( {4}^{2} )$

cancel out $\log_4$ from both sides:

$\sf \displaystyle \: - 6 x + {x}^{2} = {4}^{2}$

simplify squares:

$\sf \displaystyle \: - 6 x + {x}^{2} = 16$

move left hand side expression to right hand side and change its sign:

since we are moving left hand side expression to right hand side there'll be only 0 left in the left hand side

$\sf \displaystyle \: - 6 x + {x}^{2} - 16 = 0$

rewrite it to standard form i.e ax²+bx+c=0

$\sf \displaystyle \: {x}^{2} - 6x - 16 = 0$

rewrite -6x as 2x-8x:

$\sf \displaystyle \: {x}^{2} + 2x - 8x - 16 = 0$

factor out x and 8:

$\sf \displaystyle \: x {(x}^{} + 2) - 8(x + 2) = 0$

group:

$\sf \displaystyle \: (x - 8){(x}^{} + 2) = 0$

$\displaystyle \: x = 8 \\ x = - 2$

$\therefore \: x = - 2$

<h3>Question-7:</h3>

move left hand side log to right hand side:

$\displaystyle \: \log(x ) + \log(x - 21) = 2$

use mutilation logarithm rule;

$\displaystyle \: \log( {x}^{2} - 21x) = 2$

$\log(10^2)=2$ so

$\displaystyle \: \log( {x}^{2} - 21x) = \log({10}^{2} )$

cancel out log from both sides:

$\displaystyle \: {x}^{2} - 21x = 100$

make it standard form:

$\displaystyle \: {x}^{2} - 21x - 100= 0$

factor:

$\displaystyle \: {(x} + 4)(x - 25)= 0$

so

$\displaystyle \: x = 25$

5 0

2 years ago

The original price p of an item less a discount of 20%

Otrada [13]

Like 10% maybe your subtracting ight? message back so i can help

7 0

3 years ago

Read 2 more answers