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2 years ago

Tan^-1[-squareroot3/3] what is the degree

Mathematics

2 answers:

lozanna [386]2 years ago

7 0

Answer:

π/6

Step-by-step explanation:

if tan(pi/16) = squareroot(3)/3

then tan^-1(squareroot(3)/3) = pi/6

Reika [66]2 years ago

3 0

Answer:

150˚ or 330˚

Step-by-step explanation:

It is easier to use the calculator, but you can use the unit circle too.

Calculator:

Plug in the equation without the negative, and you get 30˚ which is your reference angle. But since tangent is negative, the angle is in either the second quadrant, or the fourth (because tangent is negative in these quadrants). Now using your reference angle, you can find the answer.

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Can someone solve this for me? Common factor first! Thank you.

USPshnik [31]

Answer:

$5 {(x - 4)}^{2}$

Step-by-step explanation:

$5 {x}^{2} - 40x + 80 \\ =5( {x}^{2} - 8x + 16) \\ = 5 ( {x}^{2} - 8x + {4}^{2} ) \\ = 5 {(x - 4)}^{2}$

5 0

2 years ago

Read 2 more answers

There are 3 consecutive integers. Four times the smallest integer is equal to six more than three times the largest integer. Wha

Aliun [14]

Answer:

Step-by-step explanation:

Let the smallest integer equal x - 1, the next one equal x, and the last one equal x + 1.

4(x-1) = 3(x-1) + 6

x-1 = 5

x = 6

The smallest integer is x - 1, which is equal to 5. Thus, the smallest of the three integers is 5.

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2 years ago

A food packet is dropped from a helicopter and is modeled by the function f(x) = −15x2 + 6000. The graph below shows the height

melisa1 [442]

General Idea:

Domain of a function means the values of x which will give a DEFINED output for the function.

Applying the concept:

Given that the x represent the time in seconds, f(x) represent the height of food packet.

Time cannot be a negative value, so

$x\geq 0$

The height of the food packet cannot be a negative value, so

$f(x)\geq 0$

We need to replace $-15x^2+6000$ for f(x) in the above inequality to find the domain.

$-15x^2+6000\geq 0 \; \; [Divide \; by\; -15\; on\; both\; sides]\\ \\ \frac{-15x^2}{-15} +\frac{6000}{-15} \leq \frac{0}{-15} \\ \\ x^2-400\leq 0\;[Factoring\;on\;left\;side]\\ \\ (x+200)(x-200)\leq 0$

The possible solutions of the above inequality are given by the intervals $(-\infty , -2], [-2,2], [2,\infty )$ . We need to pick test point from each possible solution interval and check whether that test point make the inequality $(x+200)(x-200)\leq 0$ true. Only the test point from the solution interval [-200, 200] make the inequality true.