All categories

3 years ago

Find the solution set of x + y = 5, 2 x - y = 7

1 answer:

4 0

X + y = 5
x = 5 - y

2x - y = 7
2(5-y) - y = 7
10 - 2y - y = 7
10 - 3y = 7
10 - 7 = 3y
3 = 3y
y = 1

x + y = 5
x + 1 = 5
x = 5 - 1
x = 4

You might be interested in

In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a fur

Margaret [11]

Answer:

D = -4/7 = - 0.57

C = 17/7 = 2.43

Explanation:

We have the following two equations:

$3C + 4D = 5\ --------------- eqn (1)\\2C + 5D = 2\ --------------- eqn (2)$

First, we isolate C from equation (2):

$2C + 5D = 2\\2C = 2 - 5D\\C = \frac{2 - 5D}{2}\ -------------- eqn(3)$

using this value of C from equation (3) in equation (1):

$3(\frac{2-5D}{2}) + 4D = 5\\\\\frac{6-15D}{2} + 4D = 5\\\\\frac{6-15D+8D}{2} = 5\\\\6-7D = (5)(2)\\7D = 6-10\\\\D = -\frac{4}{7}$

Put this value in equation (3), we get:

$C = \frac{2-(5)(\frac{-4}{7} )}{2}\\\\C = \frac{\frac{14+20}{7}}{2}\\\\C = \frac{34}{(7)(2)}\\\\C = \frac{17}{7}\\$

5 0

3 years ago

An AC source operating at 59 Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and an induc

Alexxandr [17]

I = V/Z

V = voltage, I = current, Z = impedance

First let's find the total impedance of the circuit.

The impedance of the resistor is:

$Z_{R}$ = R

R = resistance

Given values:

R = 1200Ω

Plug in:

$Z_{R}$ = 1200Ω

The impedance of the inductor is:

$Z_{L}$ = j2πfL

f = source frequency, L = inductance

Given values:

f = 59Hz, L = 2.4H

Plug in:

$Z_{L}$ = j2π(59)(2.4) = j889.7Ω

Add up the individual impedances to get the Z, and convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$

Z = 1200 + j889.7

Z = 1494∠36.55°Ω

I = V/Z

Given values:

V = 170∠0°V (assume 0 initial phase)

Z = 1494∠36.55°Ω

I = 170∠0°/1494∠36.55°Ω

I = 0.1138∠-36.55°A

Round the magnitude of I to 2 significant figures and now you have your maximum current:

I = 0.11A

5 0

3 years ago

The amplitude of a sound wave determines its

Savatey [412]

The Loudness i believe.

5 0

3 years ago

Read 2 more answers

A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time

IrinaK [193]

Answer:

$8.0\mu C$

Explanation:

We are given that

$f=1.6 Hz$

$q=3.0\mu C=3.0\times 10^{-6} C$

$1\mu C=10^{-6} C$

Current,I= $75\mu A=75\times 10^{-6} A$

$1\mu A=10^{-6} A$

We have to find the maximum charge of the capacitor.

Charge on the capacitor, $q=q_0cos\omega t$

$\omega=2\pi f=2\pi\times 1.6=3.2\pi rad/s$

$3\times 10^{-6}=q_0cos3.2\pi t$ ....(1)

$I=\frac{dq}{dt}=-q_0\omega sin\omega t$

$75\times 10^{-6}=-q_0(3.2\pi)sin3.2\pi t$ ....(2)

Equation (2) divided by equation (1)

$-3.2\pi tan3.2\pi t=\frac{75\times 10^{-6}}{3\times 10^{-6}}=25$

$tan3.2\pi t=-\frac{25}{3.2\pi}=-2.488$

$3.2\pi t=tan^{-1}(-2.488)=-1.188rad$

$q_0=\frac{q}{cos\omega t}=\frac{3\times 10^{-6}}{cos(-1.188)}=8.0\times 10^{-6}=8\mu C$

Hence, the maximum charge of the capacitor= $8.0\mu C$

4 0

4 years ago

Explain how wind erosion changes land forms

valkas [14]

Answer:

the wind carries abrasive materials

Explanation:

such as sand and salt over time theses small particles slowly strip way at the land form sculpting it by eroding the softer layers first

8 0

2 years ago