Can someone check my answers please

What is the resistance of a 1.00X10^2 -Ω , a 2.50-kΩ , and a 4.00-k Ω resistor connected in parallel?

Andreas93 [3]

Answer:

Therefore, the total Resistance: R = 93.9 Ω

Step-by-step explanation:

Given

R₁ = 1 × 10²

R₂ = 2.5 kΩ = 2.5 × 10³ Ω

R₃ = 4 kΩ = 4 × 10³ Ω

Given that the given resisters are connected parallel, so using the formula to calculate the total Resistance R:

$\frac{1}{R}\:=\:\frac{1}{R_1}\:+\:\frac{1}{R_2}+\frac{1}{R_3}$

susbtituting R₁ = 1 × 10², R₂ = 2.5 × 10³ Ω, and R₃ = 4 × 10³ Ω

$\frac{1}{R}=\frac{1}{1\times \:10^2}+\frac{1}{2.5\times \:10^3}+\frac{1}{4\times \:10^3}$

Multiply by LCM of R, 100, 2500, and 4000: 20000 R

$\frac{1}{R}\cdot \:20000R=\frac{1}{1\cdot \:10^2}\cdot \:20000R+\frac{1}{2.5\cdot \:10^3}\cdot \:20000R+\frac{1}{4\cdot \:10^3}\cdot \:20000R$

simplify

$20000=213R$

Switch sides

$213R=20000$

Divide both sides by 213

$\frac{213R}{213}=\frac{20000}{213}$

Simplify

$R=\frac{20000}{213}$

$R = 93.9$ Ω

Therefore, the total Resistance: R = 93.9 Ω

4 0

3 years ago

After Halloween, pumpkins were 25% off. Amy bought 2 big

yKpoI14uk [10]

Yes that’s true. You got that correct everyone listen to them.

5 0

3 years ago

if There are 24 students in me hunts class if of the student are girls what is the ratio of boys to girls in his class

Iteru [2.4K]

Answer:

24:0

Step-by-step explanation:

since all the students are girls so boys are nill

6 0

3 years ago

Can someone help me out with some pesky math quiz :) I put a pic

Sindrei [870]

1. Simplifying ∛54 gives 3∛2

2. 27^1/3 * (27)^3 is equal to 27

3. The expression (2j4k)^4/5 can be written as $\sqrt[4]{32j^{20}k^{5} }^{5}$ making the first option the correct option

4. simplifying $\sqrt[4]{16x^{4} y^{12} }$ will result to 2x∛y

5. simplifying $\sqrt[5]{2940x^{13} y^{7} }$ will result to $14x^{6} y^{3} \sqrt{15xy}$ making the last option the correct option

<h3>How to simplify $\sqrt[5]{2940x^{13} y^{7} }$ </h3>

<u>given data</u>

$\sqrt[5]{2940x^{13} y^{7} }$

= ( 2940 * x^13 * y^7 )^1/5

$= ( 2^2 * 5 * 3 * 7^2 * x^{12} * x * y^6 * y )^{1/5}$

$= ( 2^2 * 5 * 3 * 7^2 * x^{12} * x * y^6 * y )^{1/5*5/2}$

$= ( 2^2 * 5 * 3 * 7^2 * x^{12} * x * y^6 * y )^{5/5*1/2}$

multiplying the values inside the parenthesis by ^1/2

$= ( 2 * 5^{1/2} * 3^{1/2} * 7 * x^{6} * x^{1/2} * y^3 * y^{1/2} )$

$= ( 14* 5^{1/2} * 3^{1/2} * x^{6} * x^{1/2} * y^3 * y^{1/2} )$

$= ( 14 * x^6 * y^3 ) ( 5^{1/2} * 3^{1/2} * x^{1/2} * y^{1/2} )$

$= ( 14x^6y^3 ) ( 15^{1/2} * x^{1/2} * y^{1/2} )$

$= ( 14x^6y^3 ) ( 15xy)^{1/2}$

$= ( 14x^6y^3 ) \sqrt{( 15xy)}$

we can therefore say that the last option is the answer

Read more on equations here: brainly.com/question/22688504

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3 0

1 year ago

A bowl contains 9 marbles with letters on them. 4 marbles have the letter R on them. 3 marbles have the letter Son

Firdavs [7]

The probability of getting an S on the first draw and then a T on the second draw is 2/12 if the bowl contains 9 marbles with letters on them.

<h3>What is probability?</h3>

It is defined as the ratio of the number of favorable outcomes to the total number of outcomes, in other words the probability is the number that shows the happening of the event.

Probability of getting letter an S:

P(S) = 3/9

Probability of getting letter an T:

P(T) = 2/8 (without replacement)

P(S, T) = (3/9)(2/8)

P(S, T) = 6/72 = 2/12

Thus, the probability of getting an S on the first draw and then a T on the second draw is 2/12 if the bowl contains 9 marbles with letters on them.

Learn more about the probability here:

brainly.com/question/11234923

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7 0

2 years ago