In a parallel circuit, ET = 120 V, R = 30 Ω, and XL = 60 Ω. What is the phase angle (angle theta)?

HELP ME PLEASE! tell me if 11,12 are right and help w/ 13

Marysya12 [62]

Answer: I know 12 is right

Step-by-step explanation:

7 0

3 years ago

Ayuda por favor

sergeinik [125]

A partir de la definición de razón y la teoría de semejanza entre triángulos, la razón del área del triángulo AMN y el área del cuadrilátero BMNC es equivalente a 1/3.

<h3>¿Cómo determinar la medida de un lado de un triángulo desconocido?</h3>

En este problema tenemos un sistema formado por dos triángulos similares, la semejanza entre los dos triángulos se debe a la colinealidad entre los segmentos de línea AP' (triángulo pequeño) y AP'' (triángulo grande), así como de los lados AM y AB, así como los lados AN y AC, así como los mismos ángulos en la misma distribución. (Semejanza Lado - Ángulo - Lado)

En consecuencia, obtenemos las siguientes proporciones:

AP'/AP'' = MN/BC = 1/2 (1)

Finalmente, la proporción entre el triángulo AMN y el cuadrilátero BMNC es:

$\frac{AMN}{ABC - AMN} = \frac{\frac{1}{2}\cdot a \cdot \left(\frac{1}{2}\cdot h \right)}{\frac{1}{2}\cdot (2\cdot a) \cdot h - \frac{1}{2}\cdot a \cdot \left(\frac{1}{2}\cdot h \right)} = \frac{\frac{1}{4}\cdot a\cdot h }{a\cdot h - \frac{1}{4}\cdot a \cdot h }$

$\frac{AMN}{ABC - AMN} = \frac{\frac{1}{4} }{\frac{3}{4} } = \frac{1}{3}$

A partir de la definición de razón y la teoría de semejanza entre triángulos, la razón del área del triángulo AMN y el área del cuadrilátero BMNC es equivalente a 1/3.

Para aprender sobre triángulos semejantes: brainly.com/question/21730013

#SPJ1

3 0

2 years ago

After John worked at a job for 10 years, his salary doubled. If he started at $x, his salary after 10 years is _____. $1/2x $x +

dlinn [17]

2X is the answer. I am sure.

3 0

3 years ago

Need this really quick plz

Oduvanchick [21]

Answer:

Let

x=sin-¹u

Sinx=u

let y=tan-¹v

tany=v

Substituting

Sin[x + y]

Applying the sine expansion

Sinxcosy + CosxSiny

Recall x =Sin-¹u

y=tan-¹v

Sin(Sin-¹u)Cos(tan-¹v) +Cos(sin-¹u)Sin(tan-¹v)

Now at this point

Here's what you do

For the first expression

Sin(Sin-¹u)

Let's simplify this

Let P = Sin-¹u

Taking sine of both sides

SinP=u

Draw a Right angled angle for this

Since Sine from SOHCAHTOA is OPP/HYP

Where P is the angle and u is the opposite and 1 is the hypotenuse since u is the same as u/1

substituting Sin-¹u = P

You have

Sin(Sin-¹u) = SinP

and from the triangle you drew

SinP = u

Taking the second express

Cos(tan-¹v)

Let Q=Tan-¹v

taking tan of both sides

tanQ=v

Draw a right angled triangle for this too

Since Tan from SOHCAHTOA is OPP/ADJ

Find the Hypotenuse cos you'll need it

Now Let's do the substitution again

We first said tan-¹v = Q

When we substitute it in Cos(tan-¹v)

We have CosQ

Cos Q from the second right angle triangle you drew is 1/√1+v²

Because CAH is adj/Hyp

So

the first part of the original Express

Which is

Sin(Sin-¹u)Cos(tan-¹v) is now simplified to

u(1/√1+v²).

Let's Move to the second part of the Original Expression

Cos(Sin-¹u)Sin(tan-¹v)

From our first solution

We said Sin-¹u= P

So replacing it here

we have Cos(sin-¹u) = CosP

let's leave the second one for now which is sin(tan-1v) We'll deal with this after the first

so Cos(Sin-¹u) = CosP

we can still use our first Right angle triangle for this because the angle was P.

so Cos P from that triangle will be

CosP= √1-u²

Now onto the next

Sin(tan-¹v)

From the Second solution of the first we did

we said let Tan-¹v =Q

Substituting this

we have

Sin(tan-¹v) = SinQ

using the second Right angle triangle because its angle is Q

We have

SinQ= v/√1+v²

Answer for second phase Which is

Cos(sin-¹u)Sin(tan-¹v) = √1-u²(v/√1+v²)

We're done

compiling our answers

The answer to

Sin(Sin-¹u - tan-¹v) = u(1/√1+v²) + [(√1-u²)(v/√1+v²)]

You can still choose to factor out 1/√1+v² since it appears on both sides

8 0

2 years ago

What is the value of h for the parallelogram?

anastassius [24]

Answer:

9.6 units

Step-by-step explanation:

The area of the parallelogram is the product of the base length and distance between parallel sides, either way you figure it.

16 × 6 = area = 10 × h

96 = 10h

h = 96/10 = 9.6 . . . . units

3 0

3 years ago