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s344n2d4d5 [400]
3 years ago
12

If R is the total resistance for a parallel circuit with two resistors of resistances of r1 and r2 then 1/r=1/r1+1/r2 find the r

esistance r1 if the total resistance is 20 ohms and r2 is 75 ohms round your answer to the nearest ohms if neccesary
a.16 ohms
b.1405 ohms
c.27 ohms
d.102 ohms
Mathematics
2 answers:
Allushta [10]3 years ago
6 0
A little Algebra, and we're home free!

We are given R = 20 and r2 = 75. We need to find r1.

The formula is 1/R = 1/r1 + 1/r2. Plug-in:
1/20 = 1/r1 + 1/75
1/20 - 1/75 = 1/r1
0.0366 = 1/r1
r1 = 1/0.03666
r1 = 27.272

Our answer is r1 = 27 ohms.
nikitadnepr [17]3 years ago
3 0

Answer:

The correct answer is c. 27\Omega

Step-by-step explanation:

- The first step is to get the r_1 variable for alone on one side.

\frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}

  • Pass the term \frac{1}{r_2} to subtract to the left side.

\frac{1}{r}-\frac{1}{r_2}=\frac{1}{r_1}

  • Pass the term \frac{1}{r}-\frac{1}{r_2} to divide to the right side and pass the term r_1 to multiply to the left side.

r_1=\frac{1}{\frac{1}{r}-\frac{1}{r_2}}

- The second step is to replace the values of r and r_2 into the previous expression.

r_1=\frac{1}{\frac{1}{20}-\frac{1}{75}}

- The third step is to perform the operations to simplify the expression.

r_1=\frac{1}{0.05-0.01333}

r_1=\frac{1}{0.03666}

r_1=27.2727\Omega

As it is necessary to round the answer, the value of r_1 is 27\Omega.

Thus, the correct answer is c. 27\Omega

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Kevin will take 4 math tests this term. All of the tests are worth the same number of points. After taking the first 3 tests, hi
Makovka662 [10]

Answer:

Kevin needs 96 points on his last test to raise his mean test score to 90 points.

Step-by-step explanation:

we know that

The mean score is the total of all scores divided by the total number of tests.

Let

x_1 ----> the score in the first math test

x_2 ----> the score in the second math test

x_3 ----> the score in the third math test

x_4 ----> the score in the fourth math test

we have

After taking the first 3 tests, his mean test score is 88 points

so

\frac{x_1+x_2+x_3}{3} =88

x_1+x_2+x_3=264 ----> equation A

How many points does he need on his last test to raise his mean test score to 90 points?

so

\frac{x_1+x_2+x_3+x_4}{4} =90

x_1+x_2+x_3+x_4=360 ----> equation B

substitute equation A in equation B

264 + x_4 = 360

solve for x_4

x_4 = 360-264

x_4 = 96

Therefore

Kevin needs 96 points on his last test to raise his mean test score to 90 points.

4 0
4 years ago
A second particle, Q, also moves along the x-axis so that its velocity for 0 £ £t 4 is given by Q t cos 0.063 ( ) t 2 v t( ) = 4
Vladimir79 [104]

Answer:

The time interval when V_Q(t) \geq 60  is at  1.866 \leq t \leq 3.519

The distance is 106.109 m

Step-by-step explanation:

The velocity of the second particle Q moving along the x-axis is :

V_{Q}(t)=45\sqrt{t} cos(0.063 \ t^2)

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.

We are also to that :

V_Q(t) \geq 60    between   0 \leq t \leq 4

The schematic free body graphical representation of the above illustration was attached in the file below and the point when V_Q(t) \geq 60  is at 4 is obtained in the parabolic curve.

So, V_Q(t) \geq 60  is at  1.866 \leq t \leq 3.519

Taking the integral of the time interval in order to determine the distance; we have:

distance = \int\limits^{3.519}_{1.866} {V_Q(t)} \, dt

= \int\limits^{3.519}_{1.866} {45\sqrt{t} cos(0.063 \ t^2)} \, dt

= By using the Scientific calculator notation;

distance = 106.109 m

4 0
3 years ago
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Alexxx [7]

Answer:

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the line

Step-by-step explanation:

we have

y>3x-8

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The slope of the dashed line is positive m=3

The y-intercept of the dashed line is -8

see the attached figure to better understand the problem

4 0
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Ymorist [56]

Answer:

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