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2 years ago

The larger of two numbers is four more than the smaller number.If the sum of the numbers is 61,find the numbers

Mathematics

1 answer:

sergij07 [2.7K]2 years ago

6 0

Answer:

<u>28.5 and 32.5</u>

Step-by-step explanation:

<em>create an equation to solve:</em>

x + (x+4) =61

2x + 4 = 61 combine like terms

2x = 57 subtract 4 from both sides

x = 28.5 divide 2 from both sides

28.5 is the smaller number

the larger number is 4 more so:

28.5 + 4 = 32.5

32.5 is the larger number

check your work:

28.5 + 32.5 = 61

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Read 2 more answers

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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3 years ago

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