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siniylev [52]
3 years ago
10

Evaluate 2r (t – 1) ifr = 4 and t = 6

Mathematics
2 answers:
Dominik [7]3 years ago
5 0

Answer:

40

Step-by-step explanation:

2r (t – 1)

Let r= 4 and t = 6

2*4 ( 6-1)

PEMDAS says parentheses first

2*4(5)

Then multiply from left to right

8 ( 5)

40

garri49 [273]3 years ago
4 0

Answer:

40

Step-by-step explanation:

Step 1: Define

2r(t - 1)

r = 4

t = 6

Step 2: Substitute and Evaluate

2(4) · (6 - 1)

8 · 5

40

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Please help with this will give brainliest to the first answer!
kozerog [31]

Answer:

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Step-by-step explanation:

ok the equation  in finding the volume of a come is

V= \pi r^2 \frac{h}{3}

step 1: plugin

\pi =3.14

3.14 x 6^2 x 11/3

3.14x 36 x 11/3 =   (6/3 turns into a 2)

113.04x 11/3= 114. 48, when rounded, it's going to give you 415 in^3

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2 years ago
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Maksim231197 [3]

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7 0
3 years ago
A candle burned at a steady rate. After 33 minutes, the candle was 11.2 inches tall. Eighteen minutes later, it was 10.75 inches
Amiraneli [1.4K]

Answer:

9.025 inches

Step-by-step explanation:

Let the length be represented by:

L = mt+L_0

Where L is the length of candle at time t

m is the rate at which the candle is burning.

And L_0 is the initial length of the candle.

As per the question statement, let us put the given values in the equation.

11.2 = m\times 33 + L_0 .... (1)\\10.75 = m\times 51+L_0 ..... (2)

Subtracting (1) from (2):

0.45 = -18m\\\Rightarrow m = -0.025

Putting the value of m in the equation (1):

11.2 = -0.025 \times 33+L_0\\\Rightarrow L_0 = 12.025

Therefore, the equation becomes:

L = -0.025t + 12.025

Now, we have to find the height of candle after 2 hours.

2 hours mean 120 minutes.

L = -0.025 \times 120 + 12.025\\\Rightarrow L = 9.025\ inches

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3 0
2 years ago
Describe and correct the error made in solving the system of equations by elimination
lesya692 [45]
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correction :
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-x + 2y + 3z = 10...multiply by 5
---------------------
5x + 3y - z = 15
-5x + 10y + 15z = 50 (result of multiplying by 5)
--------------------add
13y + 14z = 65

6 0
3 years ago
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