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3 years ago

Solve for q; 6n-5q/11t = c

Mathematics

1 answer:

Ghella [55]3 years ago

7 0

Answer:

$\large\boxed{q=\dfrac{6n-11ct}{5}}$

Step-by-step explanation:

$\dfrac{6n-5q}{11t}=c\qquad\text{multiply both sides by}\ 11t\neq0\\\\6n-5q=11ct\qquad\text{subtract}\ 6n\ \text{from both sides}\\\\-5q=11ct-6n\qquad\text{change the signs}\\\\5q=6n-11ct\qquad\text{divide both sides by 5}\\\\q=\dfrac{6n-11ct}{5}$

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Step-by-step explanation:

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Read 2 more answers

In a certain town, the amount of sulfur oxide in the air, S, in tons, is related to the town’s population, P, in people. The rel

Lady bird [3.3K]

Answer:

Change in sulfur oxide in the air = $\frac{\textup{110010}}{\textup{S}}$

Step-by-step explanation:

Data provided in the question:

Relation between the amount of sulfur oxide in the air and the population as:

S² = 110P² + 20P + 600

Population growth rate, $\frac{\textup{dP}}{\textup{dt}}$ = 10 people per month

Now,

change in sulfur oxide with time i.e $\frac{\textup{dS}}{\textup{dt}}$

differentiating the given relation with respect to time 't'

we have

$2S\frac{\textup{dS}}{\textup{dt}}$ = $2\times110P\frac{\textup{dP}}{\textup{dt}}$ + 20

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we have

$2S\frac{\textup{dS}}{\textup{dt}}$ = 2 × 110 × 100 × 10 + 20

$2S\frac{\textup{dS}}{\textup{dt}}$ = 220020

$\frac{\textup{dS}}{\textup{dt}}$ = $\frac{\textup{220020}}{\textup{2S}}$

Change in sulfur oxide in the air = $\frac{\textup{110010}}{\textup{S}}$

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2 years ago