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3 years ago

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In a distribution system that's operating at 4.160 kV, what is the voltage level in volts? A. 4.160 volts B. 4,160 volts C. 0.00

4160 volts D. 4,160,000 volts

1 answer:

Romashka-Z-Leto [24]3 years ago

4 0

The answer to this question is a 4160 volta

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Are rhesw ratios equivalent 6:2 and 7:3

morpeh [17]

Answer:

No, they are not equivalent.

Step-by-step explanation:

There are two ways to check this.

1. You can multiply the means and extremes. If the ratios are equivalent, they products of the means and extremes will be equal.

Multiply the means (2 and 7) and multiply the extremes (6 and 3)

14 ≠ 18

2. You could also check these by dividing them.

6/2 = 7/3

3 ≠ 2.3

Since neither one of the checks are correct, 6:2 and 7:3 are not equivalent.

I really hope this helps =)

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3 years ago

Solve the equation for c.

Ede4ka [16]

Answer = 8

Because:

c(c + 3) - c(c-4) = 9c - 16
c^2 + 3c - c^2 + 4c = 9c - 16
3c + 4c = 9c - 16
7c = 9c - 16
7c + 16 = 9c
16 = 2c
8 = c

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3 years ago

Antoinette solves the linear equation 3(x-3)+2x+9=2x+2(x-1)

xeze [42]

Hope this worked :3

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4 years ago

Use the augmented matrix method to solve the following system of equations. Your answers may be given as decimals or fractions.

Alexandra [31]

Answer:

$x\ =\ \dfrac{21}{5}$

$y\ =\ \dfrac{3}{5}$

z = 2

Step-by-step explanation:

Given equations are

x - 2y - z = 2

x + 3y - 2z = 4

-x + 2y + 3z = 2

from the given equations the augmented matrix can be written as

$\left[\begin{array}{ccc}1&-2&-1:2\\1&3&-2:4\\-1&2&3:2\end{array}\right]$

$R_2=>R_2-R_1\ and\ R_3=>R_3+R_1$

$=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&5&-1:2\\0&0&2:4\end{array}\right]$

$R_2=>\dfrac{R_2}{5}$

$=\ \left[\begin{array}{ccc}1&-2&-1:2\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\0&0&2:4\end{array}\right]$

$R_1=>R_1+2.R_2$

$=\ \left[\begin{array}{ccc}1&0&-1-\dfrac{2}{5}:2+\dfrac{4}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right]$

$=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&2:4\end{array}\right]$

$R_3=>\dfrac{R_3}{2}$

$=\ \left[\begin{array}{ccc}1&0&\dfrac{-7}{5}:\dfrac{14}{5}\\\\0&1&\dfrac{-1}{5}:\dfrac{2}{5}\\\\0&0&1:2\end{array}\right]$

$R_1=>R_1+\dfrac{7}{5}R_3\ and\ R_2+\dfrac{1}{5}R_3$

$=\ \left[\begin{array}{ccc}1&0&0:\dfrac{14}{5}+\dfrac{7}{5}\\\\0&1&0:\dfrac{2}{5}+\dfrac{1}{5}\\\\0&0&1:2\end{array}\right]$

So, from the above augmented matrix, we can write

$x\ =\ \dfrac{21}{5}$

$y\ =\ \dfrac{3}{5}$

z = 2

7 0

3 years ago

F(t)=-3t+4 f(__)=19

tangare [24]

Answer:

f(-5)=19, so t= -5

Step-by-step explanation:

We begin with the function f(t). To find when it equals 19, we can set the function equal to 19 and then solve for t

$f(x)=-3t+4\\-3t+4=19\\\\-3t=15\\\\t=-5$

7 0

3 years ago