All categories

2 years ago

Ttm question need help

Mathematics

1 answer:

azamat2 years ago

7 0

Answer:

The first option is the correct answer

Step-by-step explanation:

You might be interested in

En un triángulo rectángulo A es un ángulo agudo y Sen A = 4/5 ¿Cuál será el valor de Tan A?

Nonamiya [84]

Answer:

$\displaystyle \tan A=\frac{4}{3}$

Step-by-step explanation:

Funciones Trigonométricas

La identidad principal en trigonometría es:

$sen^2A+cos^2A=1$

Si sabemos que A es un ángulo agudo (que mide menos de 90°), su seno y coseno son positivos.

Dado que Sen A = 4/5, calculamos el coseno:

$cos^2A=1-sen^2A$

Sustituyendo:

$\displaystyle cos^2A=1-\left(\frac{4}{5}\right)^2$

$\displaystyle cos^2A=1-\frac{16}{25}$

$\displaystyle cos^2A=\frac{25-16}{25}$

$\displaystyle cos^2A=\frac{9}{25}$

Tomando raíz cuadrada:

$\displaystyle cos\ A=\sqrt{\frac{9}{25}}=\frac{3}{5}$

La tangente se define como:

$\displaystyle \tan A=\frac{sen\ A}{cos\ A}$

Substituyendo:

$\displaystyle \tan A=\frac{\frac{4}{5}}{\frac{3}{5}}$

$\displaystyle \tan A=\frac{4}{3}$

6 0

2 years ago

Plz help meeeeeeee I need to turn this in today plzzz!!!!!!

Katarina [22]

Answer:

Step-by-step explanation:

Step 1 Answer: Yes, 3

Step 2 Answer: Yes, 5

Step 3 Answer: No

8 0

2 years ago

Read 2 more answers

-45×68 what is the answer

Marianna [84]

Answer: -3060

68
-45
--------
340
272x <-- the x is in place of a zero
--------
3060

times the negative: -3060

4 0

2 years ago

State the slope of the line. A:undefined B:1 C:-2 D:0 Help ASAP

Nikitich [7]

Every point on this line has the same y coordinate, which is -2. There is no "rise" at all so rise = 0 while the "run" is any nonzero value you want. If you made the run be say 2, then

slope = rise/run = 0/2 = 0

The slope is 0

Answer: D) 0

note: all horizontal lines like this have a slope of 0. It doesn't matter what the y intercept is

8 0

3 years ago

Read 2 more answers

If cos(θ)=2853 with θin Q IV, what is sin(θ)?

marysya [2.9K]

Answer: $\sin \theta=\frac{-45}{53}$

Step-by-step explanation:

Since we have given that

$\cos\theta=\frac{28}{53}$

And we know that θ is in the Fourth Quadrant.

So, Except cosθ and sec θ, all trigonometric ratios will be negative.

As we know the "Trigonometric Identity":

$\cos^2\theta+\sin^2\theta=1\\\\\sin \theta=\sqrt{1-\cos^2\theta}\\\\\sin \theta=\sqrt{1-(\frac{28}{53})^2}=\sqrt{\frac{53^2-28^2}{53^2}}\\\\\sin \theta=\sqrt{\frac{2025}{53^2}}\\\\\sin \theta=\frac{45}{53}$

It must be negative due to its presence in Fourth quadrant.

Hence, $\sin \theta=\frac{-45}{53}$

7 0

3 years ago