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3 years ago

If 0° ≤ θ ≤ 90° and cosθ = 11 15 , what is the value of sin (90° - θ)?

2 answers:

7 0

Since the angles θ and 90° - θ are complementary, the cosine of one of the angles is equal to the sine of the other. Therefore, because cosθ = $\frac{11}{15}$ , the value of sin (90° - θ) must be $\frac{11}{15}$ .

kolezko [41]3 years ago

5 0

The same value as cosα

cosα=sin(90-α) and vice versa

sinα=cos(90-α)

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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>

4 0

3 years ago

Can I get some help ?

Mashcka [7]

#1
avatar
0
use the formula V*t=x so if the velocity is 8 2/3 miles per hour and you need to travel 5 1/5 miles then (8 2/3)*t=5 1/5,
so t=(5 1/5)/(8 2/3)= .6 hours
Guest Oct 30, 2016

5 0

3 years ago