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Ksju [112]
3 years ago
9

Simplify the following expression.

Mathematics
2 answers:
Murljashka [212]3 years ago
8 0

12 - 5[(4 +2)² - (22 + 10)]

Remember to follow PEMDAS. First, solve both parenthesis

12 - 5[(4 +2)² - (22 + 10)]

12 - 5[(6)² - (32)]

Next, solve the exponent

(6)² = 6 x 6 = 36

12 - 5[(36) - (32)]

Then, solve the bracket.

36 - 32 = 4

12 - 5[4]

Next, multiply

-5 x 4 = -20

Finally, subtract.

12 - 20 = -8

None of the above. The answer is -8. Please check the answers and the question again

<em>~Rise Above the Ordinary</em>


adoni [48]3 years ago
7 0
$$ 12 - 5[(4 + 2)\div\2 - (22 + 10)]= 157 $$
238 this answer
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Initially, there are 40 grams of A and 50 grams of B, and for each gram of B, 2 grams of A is used. It is observed that 15 grams
hram777 [196]

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Step-by-step explanation:

let X(t) denote grams of C formed in  t mins.

For X grams of C we have:

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Amounts of A,B remaining at any given time is expressed as:

40-\frac{2}{3}Xg of A and  50-\frac{1}{3}Xg  of B

Rate at which C is formed satisfies:

\frac{dX}{dt} \infty(40-\frac{2}{3}X)(50-\frac{1}{3}X)->\frac{dX}{dt}=k(90-X)\\\therefore \frac{dX}{(90-X)^2}=kdt->\int{\frac{dX}{(90-X)^2}} \, =\int {k} \, dt  \\\therefore \frac{1}{90-X}=kt+c->90-X=\frac{1}{kt+c}\\\\X(t)=90-\frac{1}{kt+c}

Apply the initial condition,X(0)=0 ,to the expression above

0=90-\frac{1}{c} \ \ ->c=\frac{1}{90}\\\therefore\\X(t)=90-\frac{1}{kt+\frac{1}{90}} \ \ ->X(t)=90-\frac{90}{90kt+c}

Now at X(8)=15:

15=90-\frac{90}{90\times 8k+1}  \ ->75=\frac{90}{720k+1}\\k=0.0002778

Substitute  in X(t) to get

X(t)=90-\frac{90}{0.0002778t\times 90+1}\\X(t)=90-\frac{90}{0.25t+1}\\But \ t=16\\\therefore X(t)=90-\frac{90}{0.025\times16+1}\\X(t)=25.71

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3 years ago
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