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3 years ago

Solve the equations X^2+y^2 =16 x+y=1

Mathematics

1 answer:

Leviafan [203]3 years ago

5 0

Answer:

x = 1-y/6 and y= - 35/ - 37

Step-by-step explanation:

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In a parallel circuit, ET = 240 V, R = 47 kΩ, and XL = 50 kΩ. What is the reactive power?

Fittoniya [83]

Answer:

Reactive power is 1.152 VARs

Step-by-step explanation:

We are given that in a parallel circuit, $E_{T}=240V$ ,R=47kΩ and $X_{L}$ =50kΩ.

reactive power means the dissipating power resulting from inductive and capacitive loads measured in volt-amperes reactive(VAR).Since our circuit has inductor only, reactive power will be power dissipated from inductor.

Current through inductor= $\frac{E_{T}}{X_{L}} =\frac{240}{50}=4.8$ milli amps=0.0048 A

Reactive power = (current flowing through inductor)X (Voltage across inductor)

Hence reactive power = 0.0048 X 240=1.152 VARs

7 0

4 years ago

G If f(x) = 6x – 4, what is f(x) when x = 8?

I am Lyosha [343]

Answer: f(x)=44

Step-by-step explanation:

Since x is given, we can directly plug it into our function.

$f(8)=6(8)-4$

$f(8)=48-4$

$f(8)=44$

8 0

3 years ago

Damon has 5 dimes and some nickels in his pocket worth a total of $1.20. To find the number of nickels Damon has, a student wrot

ki77a [65]

In the student's equation, there are many problems. Nickels are worth 5 cents, represented by .05, not 50 dollars. Also, the coeffient 5 should not be placed there, it belongs with 5(.1), as we already know how many dimes there are. If we were able to make a new one, it would be, 5(.1)+.05n=1.20.

4 0

3 years ago

How could you find the y-coordinate of the midpoint of a vertical line segment with endpoints at (0,0) and (0,-12)

Oduvanchick [21]

I think it should be -6, because line segment equals 12 and it's midpoint will be 12/2 = 6, so you must go down by 6 from (0,0) point on y axis which is 0-6=-6

So the point will be (0, -6)

8 0

4 years ago

Solve the Anti derivative.

Alex Ar [27]

Answer:

$\displaystyle \int {\frac{1}{9x^2+4}} \, dx = \frac{1}{6}arctan(\frac{3x}{2}) + C$

General Formulas and Concepts:

Algebra I

Factoring

Calculus

Antiderivatives - integrals/Integration

Integration Constant C

U-Substitution

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Trig Integration: $\displaystyle \int {\frac{du}{a^2 + u^2}} = \frac{1}{a}arctan(\frac{u}{a}) + C$

Step-by-step explanation:

Step 1: Define

$\displaystyle \int {\frac{1}{9x^2 + 4}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Factor fraction denominator: $\displaystyle \int {\frac{1}{9(x^2 + \frac{4}{9})}} \, dx$
[Integral] Integration Property - Multiplied Constant: $\displaystyle \frac{1}{9} \int {\frac{1}{x^2 + \frac{4}{9}}} \, dx$

Step 3: Identify Variables

Set up u-substitution for the arctan trig integration.

$\displaystyle u = x \\ a = \frac{2}{3} \\ du = dx$

Step 4: Integrate Pt. 2

[Integral] Substitute u-du: $\displaystyle \frac{1}{9} \int {\frac{1}{u^2 + (\frac{2}{3})^2} \, du$
[Integral] Trig Integration: $\displaystyle \frac{1}{9}[\frac{1}{\frac{2}{3}}arctan(\frac{u}{\frac{2}{3}})] + C$
[Integral] Simplify: $\displaystyle \frac{1}{9}[\frac{3}{2}arctan(\frac{3u}{2})] + C$
[integral] Multiply: $\displaystyle \frac{1}{6}arctan(\frac{3u}{2}) + C$
[Integral] Back-Substitute: $\displaystyle \frac{1}{6}arctan(\frac{3x}{2}) + C$

Topic: AP Calculus AB

Unit: Integrals - Arctrig

Book: College Calculus 10e

7 0

3 years ago