All categories

4 years ago

Tan(theta)=40/58 How do u solve? How do u isolate theta? show work plz!

Mathematics

2 answers:

zzz [600]4 years ago

8 0

Answer:

34.60°

Step-by-step explanation:

$\tan \theta = \frac {40}{58}$

$\tan \theta = 0.689655172$

$\theta =\tan^{-1} 0.689655172$

$\theta =34.592288671\degree$

$\theta =34.60\degree$

Aneli [31]4 years ago

4 0

Answer:

$\Large \boxed{\theta = 34.59228868\°}$

Step-by-step explanation:

tan(θ) = 40/58

To isolate theta (unknown angle), we must take the inverse tangent of both the sides.

tan^-1(tan(θ)) = tan^-1(40/58)

The inverse tangent and tangent cancels.

Simplify the equation.

θ = tan^-1(20/29)

θ = 34.59228868°

You might be interested in

Write an equation that is perpendicular to y= 1/2x+5 and passes through (4,8)

goblinko [34]

Answer:

y = -2x + 16.

Step-by-step explanation:

The slope of the perpendicular line = -1 / slope of the given line

= -1 / 1/2 = -2.

Using the point slope form of the equation of a straight line:

y - y1 = m (x - x1)

y - 8 = -2(x - 4)

y - 8 = -2x + 8

y = -2x + 16 is the required equation.

4 0

4 years ago

Convert from scientific notation to standard form 4.83 x 10^-3

ANTONII [103]

4.83 x 10^-3 can be re-written in standard form by moving the decimal point 3 units to the left: 0.00483.

Note: If you want to be mathematically correct, please write 10^(-3), not 10^-3. If the exponent is +, you don't need the parentheses.

5 0

3 years ago

What’s the slope of (-3, 6), (2, 6)?

IRINA_888 [86]

Answer:

Step-by-step explanation:

(y2-y1)/(x2-x1)

6(y2)-6(y1)= 0

2(x2)- -3(x1)=5

0/5=0

4 0

3 years ago

Determine the domain and the range for this relationship.

Vladimir [108]

Answer:

domain 0 ≤ y ≤ 72 and the range is 0 ≤ y ≤ 6

Step-by-step explanation:

I hope this helps you.

7 0

3 years ago

Write the point-slope form of the equation of the line that passes through the points (-3, 5) and (-1, 4). Please answer quickly

Oksanka [162]

The point-slope form of the equation of a line is

$y - y_1 = m(x - x_1)$

where

$(x_1, y_1)$

is a point on the line, and

$m$

is the slope of the line.

We can use either one of the two given points as the point on the line.
We also need to find the slope. We can use the coordinates of the two given points to find the slope of the line.

The slope of the line that passes through points

$(x_1, y_1)$ and $(x_2, y_2)$

is

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

Let's find the slope using (-3, 5) as point 1 and (-1, 4) as point 2.

$m = \dfrac{4 - 5}{-1 - (-3)} = \dfrac{-1}{-1 + 3} = \dfrac{-1}{2} = -\dfrac{1}{2}$

Now we use the point-slope formula with point 1 and the slope we found just above.

$y - y_1 = m(x - x_1)$

$y - 5 = -\dfrac{1}{2}(x - (-3))$

4 0

3 years ago