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Alenkinab [10]
3 years ago
6

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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The function y=60+25sin(pi/6)t, where t is in months and t=0 corresponds to April 15, models the average high temperature in deg
BartSMP [9]

Answer:

The maximum temperature will be of 85 degrees, on July 15.

Step-by-step explanation:

Sine function:

The sine function oscilates between -1 and 1, and it's maximum value is:

\sin{(\frac{\pi}{2})} = 1

y=60+25sin(pi/6)t

The maximum value will occur when \sin{(\frac{\pi t}{6}}) = 1, and it will be of 60 + 25 = 85 degrees.

When will it occur?

First we find the value of t for which the value inside the function sine is \frac{\pi}{2}. So

\frac{\pi t}{6} = \frac{\pi}{2}

\frac{t}{6} = \frac{1}{2}

2t = 6

t = \frac{6}{2} = 3

That is the number of months after April 15, which is 3 months. So July 15.

8 0
3 years ago
Solve the following equation algebraically X^2=36
lara [203]

Answer:

I think the answer is C.

6 0
3 years ago
Jermaine, Keira, and Leon are buying supplies for an art class. Jermaine bought 2 canvases, 4 tubes of paint, and 2 paint brushe
maria [59]

Answer:

canvas $3

tube of paint $5

paint brush $4

Step-by-step explanation:

We need to choose variables for the prices of the different items, translate the sentences into equations, and solve a system of equations.

1. Define variables

Let c = price of 1 canvas, t = price of 1 tube of paint, and b = price of 1 brush.

2. Translate sentences into equations

"Jermaine bought 2 canvases, 4 tubes of paint, and 2 paint brushes for $34."

2c + 4t + 2b = 34

"Keira spent $22 on 1 canvas, 3 tubes of paint, and 1 paint brush."

c + 3t + b = 22

"Leon already has plenty of paint, so he bought 3 canvases and 2 paint brushes for $17."

3c + 2b = 17

The system of equations is

2c + 4t + 2b = 34       Eq. 1

c + 3t + b = 22           Eq. 2

3c + 2b = 17               Eq. 3

Since the third equation has only the variables c and b, we use the first equations to eliminate variable t and give us an equation in only c and b.

3 * Eq. 1 - 4 * Eq. 2

2c + 2b = 14

c + b = 7            Eq. 4

Eq. 3 and Eq. 4 form a system of equations in two variables.

3c + 2b = 17      Eq. 3

c + b = 7          Eq. 4

Solve Eq. 4 for b.

b = 7 - c        Eq. 5

Substitute into Eq. 3.

3c + 2(7 - c) = 17

3c + 14 - 2c = 17

c = 3

Plug in c = 3 into Eq. 5.

b = 7 - 3

b = 4

Plug in c = 3 and b = 3 into Eq. 2.

c + 3t + b = 22           Eq. 2

3 + 3t + 4 = 22

3t + 7 = 22

3t = 15

t = 5

Answer:

canvas $3

tube of paint $5

paint brush $4

7 0
3 years ago
1/3(b+4)= 1/2<br> What is the value of b?
Leni [432]

Answer:

A is 1 and b is 12

Step-by-step explanation:

4 0
3 years ago
A company sells cookies in 250-gram packs. When a particular batch of 1,000 packs was weighed, the mean weight per pack was 255
AURORKA [14]
Let X be a random variable representing the weight of a pack of cookies.
P(X < 250) = P(z < (250 - 255)/2.5) = P(z < -5/2.5) = P(z < -2) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275 = 2.3%

Therefore, we conclude that about 2.3% of the packs weighed less than 250 grams.
7 0
3 years ago
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