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2 years ago

Rt=R1 R2__, determined Rt when R1=100 and R2 =200?R1+R2 (electronics)

Mathematics

1 answer:

zloy xaker [14]2 years ago

7 0

Answer:

$R_t=300\ \Omega$

Step-by-step explanation:

Given that,

R₁ = 100 ohms

R₂ = 200 ohms

We need to find the equivalent resistance of the circuit. So,

$R_t=R_1+R_2\\\\R_t=100+200\\\\R_t=300\ \Omega$

So, the equivalent resistance is $300\ \Omega$ .

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Find the 5th term of the sequence in which T1=8 and Tn=3t n-1

yan [13]

T_n = 3 * T_(n-1)

Long way (always works!)
T_5 = 3*T_4,
T_4 = 3*T_3
T_3 = 3*T_2
T_2 = 3*T_1

T_5 = 3*3*3*3*T_1 = 81*T_1 = 81*8 = 648!

Short way (sometimes it works!)

T_n = 3^(n-1) * T_1 (this case is a geometric series of ratio-=3)
T_5 = 3^4*8 = 648

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3 years ago

I need help asap guys please I’ll give brainlest to help you out

Maru [420]

Answer:

Table c mate

Step-by-step explanation:

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3 years ago

Read 2 more answers

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Factor −25x−25

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=−25(x+1)

Answer:

−25(x+1)

7 0

3 years ago

What are the solutions to the equation

frosja888 [35]

Answer:

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

Step-by-step explanation:

You have the quadratic function $2x^2-x+1=0$ to find the solutions for this equation we are going to use Bhaskara's Formula.

For the quadratic functions $ax^2+bx+c=0$ with $a\neq 0$ the Bhaskara's Formula is:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}$

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}$

It usually has two solutions.

Then we have $2x^2-x+1=0$ where a=2, b=-1 and c=1. Applying the formula:

$x_1=\frac{-b+\sqrt{b^2-4.a.c} }{2.a}\\\\x_1=\frac{-(-1)+\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_1=\frac{1+\sqrt{1-8} }{4}\\\\x_1=\frac{1+\sqrt{-7} }{4}\\\\x_1=\frac{1+\sqrt{(-1).7} }{4}\\x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}$

Observation: $\sqrt{-1}=i$

$x_1=\frac{1+\sqrt{-1}.\sqrt{7}}{4}\\\\x_1=\frac{1+i.\sqrt{7}}{4}\\\\x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$

And,

$x_2=\frac{-b-\sqrt{b^2-4.a.c} }{2.a}\\\\x_2=\frac{-(-1)-\sqrt{(-1)^2-4.2.1} }{2.2}\\\\x_2=\frac{1-i.\sqrt{7} }{4}\\\\x_2=\frac{1}{4}-(\frac{\sqrt{7}}{4})i$

Then the correct answer is option C.

$x_1=\frac{1}{4}+(\frac{\sqrt{7}}{4})i$ and $x_2=\frac{1}{4}-(\frac{\sqrt{7} }{4})i$

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3 years ago

What is the area of a circle with a radius of 7 cm2 (Use 3.14 for

lyudmila [28]

Answer: 153.9

Step-by-step explanation:

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3 years ago

Rt=R1 R2__, determined Rt when R1=100 and R2 =200?R1+R2 (electronics)​

Rt=R1 R2__, determined Rt when R1=100 and R2 =200?R1+R2 (electronics)