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2 years ago

5 less than 3 times a number is 7. How would I write this as an equation

2 answers:

5 0

You want to break this down.

“5 less than” means you’ll subtract 5 from something.

That something is “3 times a number.” “3 times a number” means 3•x or 3x, since we’ll use x for that unknown number.

So, “5 less than 3 times a number” means “subtract 5 from 3x”: 3x-5

“Is” means “equals”. So 3x-5= something.

7, well, that’s 7.

So “5 less than 3 times a number is 7” means ”subtract 5 from 3x and set that equal to 7.”

3x-5=7

ValentinkaMS [17]2 years ago

3 0

Answer:

The equation would be 5-3x=7

Step-by-step explanation:

5 less than indicates subtraction, and 3 times a number indicates both values being multiplied with each other, so it's 3x. Hope that helped.

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2 years ago

Find the limit, if it exists. (If an answer does not exist, enter DNE.) lim (x, y) → (0, 0) x4 − 34y2 x2 + 17y2

HACTEHA [7]

Answer:

Step-by-step explanation:

Given the limit of the function $\lim_{(x,y) \to (0,0)} \frac{x^4-34y^2}{x^2+17y^2}$ , to find the limit, the following steps must be taken.

Step 1: Substitute the limit at x = 0 and y = 0 into the function

$= \lim_{(x,y) \to (0,0)} \frac{x^4-34y^2}{x^2+17y^2}\\= \frac{0^4-34(0)^2}{0^2+17(0)^2}\\= \frac{0}{0} (indeterminate)$

Step 2: Substitute y = mx int o the function and simplify

$= \lim_{(x,mx) \to (0,0)} \frac{x^4-34(mx)^2}{x^2+17(mx)^2}\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^4-34m^2x^2}{x^2+17m^2x^2}\\\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^2(x^2-34m^2)}{x^2(1+17m^2)}\\\\\\= \lim_{(x,mx) \to (0,0)} \frac{x^2-34m^2}{1+17m^2}\\$

$= \frac{0^2-34m^2}{1+17m^2}\\\\= \frac{34m^2}{1+17m^2}\\\\$

<em>Since there are still variable 'm' in the resulting function, this shows that the limit of the function does not exist, Hence, the function DNE</em>

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