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3 years ago

How much water

Mathematics

1 answer:

Romashka [77]3 years ago

6 0

I think it’s 7in and I’m dumb so idk hahaha

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A. If 3x + 10º and 4x - 5° are the angle of a linear pair find their measurement.

kaheart [24]

Answer:

85° and 95°

Step-by-step explanation:

A linear pair of angles sum to 180° , then

3x + 10 + 4x - 5 = 180

7x + 5 = 180 ( subtract 5 from both sides )

7x = 175 ( divide both sides by 7 )

x = 25

Then

3x + 10 = 3(25) + 10 = 75 + 10 = 85°

4x - 5 = 4(25) - 5 = 100 - 5 = 95°

7 0

3 years ago

Solve for t: 3x8= t+15

hodyreva [135]

Hey there!

I'm going to set up the equation slightly different.

$(3)(8)=t+5\\24=t+5\\t+15=24\\t+15-15=24-15\\t = 9 \checkmark$

8 0

3 years ago

Read 2 more answers

Solve the equation. You will not get credit if you do not show your work or 4 points

Feliz [49]

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5 0

3 years ago

1948 Olympics

mash [69]

Answer:

For 1948 Men's :Interquartile range is 1.5, Median is 58.3

For 2012 Men's: Interquartile range is 0.315, Median is 47.86

You can infer that in 2012 the swimmers were better and it was more competitive as the interquartile range was lower and the median was also lower as well

Step-by-step explanation:

1948 Men's 100m

57.3, 57.8, 58.1, 58.3, 58.3, 59.3, 59.6,1:00.5

⬆ ⬆ ⬆

Low IQR Median High IQR

Low IQR is average of 57.8 and 58.1 = 57.95

High IQR is average of 59.3 and 59.6 = 59.45

Median is average of 58.3 and 58.3 = 58.3

Interquartile range is High IQR- Low IQR

59.45 - 57.95=1.5

2012 Men's 100m

47.52, 47.53, 47.8, 47.84, 47.88, 47.92, 48.04, 48.44

⬆ ⬆ ⬆

Low IQR Median High IQR

Low IQR is average of 47.53 and 47.8 = 47.665

High IQR is average of 48.04 and 47.92 = 47.98

Median is average of 47.88 and 47.84 = 47.86

Interquartile range is High IQR-Low IQR

47.98-47.665=0.315

Read more on Brainly.com - brainly.com/question/14915771#readmore

8 0

4 years ago

A radioactive substance has a continuous decay rate of 0.056 per minute. How many grams of a 120 gram sample will remain radioac

sammy [17]

Answer:

The mass of the radioactive sample after 40 minutes is 12.8 g.

Step-by-step explanation:

The mass of the sample can be found by using the exponential decay equation:

$N_{t} = N_{0}e^{-\lambda t}$

Where:

N(t): is the amount of the sample at time t =?

N₀: is the initial quantity of the sample = 120 g

t = 40 min

λ: is the decay constant = 0.056 min⁻¹

Hence, the mass of the sample after 40 min is:

$N_{t} = N_{0}e^{-\lambda t} = 120g*e^{-0.056min^{-1}*40 min} = 12.8 g$

Therefore, the mass of the radioactive sample after 40 minutes is 12.8 g.

I hope it helps you!

5 0

3 years ago