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3 years ago

Solve for the variable R C=1/8Rt

Mathematics

2 answers:

Alex777 [14]3 years ago

8 0

C= $\frac{1}{8}$ Rt= $\frac{Rt}{8}$

Multiply both sides by 8.
8C=Rt

Divide both sides by t.
$\frac{8C}{t}$ =R

Margaret [11]3 years ago

7 0

$C=\frac{1}{8}*R*t \\\\ C=\frac{Rt}{8} \\\\ \boxed{R=\frac{8C}{t}}$

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2 years ago

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3 years ago

25 POINTS AVAILABLE

myrzilka [38]

Answer:

$\large\boxed{1.\ (-3, 0),\ r = 3}\\\boxed{2.\ (x+4)^2+(y-3)^2=36}\\\boxed{3.\ (x-2)^2+(y-1)^2=(\sqrt{34})^2}$

Step-by-step explanation:

The equation of a circle in standard form:

$(x-h)^2+(y-k)^2=r^2$

(h, k) - center

r - radius

1. We have the equation:

$(x+3)^2+y^2=9\\\\(x-(-3))^2+(y-0)^2=3^2$

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2. We have the center (-4, 3) and the radius r = 6. Substitute:

$(x-(-4))^2+(y-3)^2=6^2\\\\(x+4)^2+(y-3)^2=36$

3. We have the endpoints of the diameter: (-1, 6) and (5, -4).

Midpoint of diameter is a center of a circle.

The formula of a midpoint:

$\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

Substitute:

$h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1$

The center is in (2, 1).

The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute the coordinates of the points (2, 1) and (5, -4):

$r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}$

Finally we have:

$(x-2)^2+(y-1)^2=(\sqrt{34})^2$

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3 years ago

If y = -10 -x, find y when x = -2

arlik [135]

Answer:

y = -8

Step-by-step explanation:

y = -10 - x

y = -10 - (-2)

y = -10 + 2

y = -8

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3 years ago

Read 2 more answers

Use substitution to solve the following system of equations. What is the value of y? {3x+2y=12{5x−y=7 A) y = -3B) y = 3C) y = -2

Vlad [161]

<em>Answer</em>

B) y = 3

<em>Step-by-step explanation</em>

Given the system of equations:

$\begin{gathered} 3x+2y=12\text{ \lparen eq. 1\rparen} \\ 5x-y=7\text{ \lparen eq. 2\rparen} \end{gathered}$

Isolating x from equation 1:

$\begin{gathered} 3x+2y-2y=12-2y \\ 3x=12-2y \\ \frac{3x}{3}=\frac{12-2y}{3} \\ x=\frac{12}{3}-\frac{2}{3}y \\ x=4-\frac{2}{3}y\text{ \lparen eq. 3\rparen} \end{gathered}$

Substituting equation 3 into equation 2 and solving for y:

$\begin{gathered} 5(4-\frac{2}{3}y)-y=7 \\ 5\cdot4-5\cdot\frac{2}{3}y-y=7 \\ 20-\frac{10}{3}y-y=7 \\ 20-\frac{13}{3}y=7 \\ 20-\frac{13}{3}y-20=7-20 \\ -\frac{13}{3}y=-13 \\ (-\frac{3}{13})\cdot-\frac{13}{3}y=(-\frac{3}{13})\cdot-13 \\ y=3 \end{gathered}$

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1 year ago