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3 years ago

Greatest common factor of 21,30,44

Mathematics

1 answer:

lana [24]3 years ago

3 0

Answer:

Step-by-step explanation:

We found the factors 21,30,44 . The biggest common factor number is the GCF number. So the greatest common factor 21,30,44 is 1.

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Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz helpppppppppppppppp<br><br><br>-3 1/5 ÷2 1/5

Alla [95]

Answer:

−16×5 / 5×11

= -16/11

4 0

3 years ago

when the driver is in the empty truck, the mass is 2948.35 kilogram. the mass of 1 box of paper is 22.5 kilograms. the driver de

marishachu [46]

Answer:

0 boxes minimum

Step-by-step explanation:

The mass of the truck and paper must satisfy ...

22.5b + 2948.35 ≤ 4700 . . . . total truck mass cannot exceed bridge limits

22.5b ≤ 1751.65

b ≤ 77.85

The driver can take a minimum of 0 boxes and a maximum of 77 boxes of paper over the bridge.

_____

The question asks for the <em>minimum</em>. We usually expect such a question to ask for the <em>maximum</em>.

7 0

3 years ago

Pregnant women metabolize some drugs at a slower rate than the rest of the population. The half-life of caffeine is about 4 hour

UkoKoshka [18]

Answer:

Husband:

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Woman:

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

Step-by-step explanation:

The amount of caffeine in the body can be modeled by the following equation:

$C(t) = C(0)e^{rt}$

In which C(t) is the amount of caffeine t hours after 8 am, C(0) is how much coffee they took and r is the rate the the amount of caffeine decreases in their bodies.

110 mg of caffeine at 8 am,

So $C(0) = 110$

Husband

Half life of 4 hours. So

$C(4) = 0.5C(0) = 0.5*110 = 55$

$C(t) = C(0)e^{rt}$

$55 = 110e^{4r}$

$e^{4r} = 0.5$

Applying ln to both sides

$\ln{e^{4r}} = \ln{0.5}$

$4r = \ln{0.5}$

$r = \frac{\ln{0.5}}{4}$

$r = -0.1733$

So for the husband

$C(t) = 110e^{-0.1733t}$

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

$C(t) = 110e^{-0.1733t}$

$C(11) = 110e^{-0.1733*11} = 16.35$

The husband will have 16.35 mg of caffeine in his body at 7 pm.

Pregnant woman

Half life of 10 hours. So

$C(10) = 0.5C(0) = 0.5*110 = 55$

$C(t) = C(0)e^{rt}$

$55 = 110e^{10}$

$e^{10r} = 0.5$

Applying ln to both sides

$\ln{e^{10r}} = \ln{0.5}$

$10r = \ln{0.5}$

$r = \frac{\ln{0.5}}{10}$

$r = -0.0693$

At 7 pm

7 pm is 11 hours after 8 am, so this is C(11)

$C(t) = 110e^{-0.0693t}$

$C(11) = 110e^{-0.0693*11} = 51.33$

The pregnant woman will have 51.33 mg of caffeine in her body at 7 pm.

7 0

4 years ago

A triangle has one 40° angle and two 70° angles. Is this triangle equilateral?

Mila [183]

Answer:

Step-by-step explanation:

All angles have to be =

5 0

3 years ago

Daniela recorded the low temperatures during the school day last week and this week. Her results are shown in the table below.

mixas84 [53]

Based on the information given about the temperature, Daniela made her error on step 3 because it says: "Find the ratio of the differences of the means compared to the mean absolute deviation.

<h3>How to illustrate the error?</h3>

From the information given, it van be seen that Daniela recorded the low temperatures during the school day last week and this week and her results are shown in the table.

Here, she didn't find the difference between the today and last week means. She just put the mean as a whole.

Therefore, Daniela made her error on step 3 because it says: "Find the ratio of the differences of the means compared to the mean absolute deviation. This was the error.

Learn more about error on:

brainly.com/question/10411345

#SPJ1

5 0

2 years ago