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3 years ago

GIVING OUT 5 POINTS JUST ANWSER THIS QUESTION 2+6 TIMES 3000001 DIVIDED BY 23 AND SUBTRACT 2 AND ADD 1

Mathematics

1 answer:

Maurinko [17]3 years ago

3 0

Answer:

$\frac{23913}{25000}$

Step-by-step explanation:

According to the Order of Operations [GEMS\BOMDAS\PEMDAS etc.], Subtraction & Addition are evaluated after Division & Multiplication:

$2 + 6 \times 3000001 \div 23 - 2 + 1 \\ \\ 2 + 782608\frac{1913}{2000} - 2 + 1 = \frac{23913}{25000}$

I am joyous to assist you anytime.

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(5 - 3)⁴ - 2(7) + 8² = ?

Veseljchak [2.6K]

Answer:

Step-by-step explanation:

(5 - 3)⁴ - 2(7) + 8²

PEMDAS

Parentheses first

(2)⁴ - 2(7) + 8²

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16 - 2(7) + 64

Multiply and divide from left to right

16 -14 +64

Add and subtract from left to right

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3 years ago

What is the value of r? 1.2r−1 1.3r=40

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3 years ago

What is the solution of log2x + 6 144 = 2?

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3 years ago

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What is the image below simplified as a fraction?

marysya [2.9K]

Answer:

159.8195397043133

Step-by-step explanation:

2²⁰

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

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

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:

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

$\displaystyle \: x = 25$

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2 years ago