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pav-90 [236]
3 years ago
9

Mr. Mole left his burrow and started digging his way down at a constant rate.

Mathematics
2 answers:
Iteru [2.4K]3 years ago
6 0

Constant rate means it is linear so we don't have to worry about fluctuation.

change in y/ change in x=-18-(-25.2)/5-8=2.4 meters per second

hammer [34]3 years ago
5 0

We have been given that Mr. Mole left his burrow and started digging his way down at a constant rate. The table compares Mr. Mole altitude relative to the ground (in meters) and the time since he left (in minutes).

Mr. Mole's speed will be equal to slope of line passing through the given points.

m=\frac{y_2-y_1}{x_2-x_1}

Let point (5,-18)=(x_1,y_1) and point (8,-25.2)=(x_2,y_2).

Upon substituting these values in slope formula, we will get:

m=\frac{-25.2-(-18)}{8-5}

m=\frac{-25.2+18}{3}

m=\frac{-7.2}{3}

m=-2.4  

Mr. Mole's altitude is getting more and more negative as he digging down.

Since speed cannot be negative, therefore, Mr. Mole's speed is 2.4 meters per minute.

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8 0
2 years ago
Please please help me
Gekata [30.6K]

Answer:

\large\boxed{\dfrac{\boxed{8}}{81}}

Step-by-step explanation:

\text{If}\ a_1,\ a_2,\ a_3,\ a_4,\ ...,\ a_n\ \text{is the geometric sequence, then}\\\\\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...=\dfrac{a_n}{a_{n-1}}=constant=r-\text{common ration}.\\\\\text{We have}\ \dfrac{1}{2},\ \dfrac{1}{3},\ \dfrac{2}{9},\ \dfrac{4}{27},\ ...\\\\\dfrac{\frac{1}{3}}{\frac{1}{2}}=\dfrac{1}{3}\cdot\dfrac{2}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{2}{9}}{\frac{1}{3}}=\dfrac{2}{9}\cdot\dfrac{3}{1}=\dfrac{2}{3}\\\\\dfrac{\frac{4}{27}}{\frac{2}{9}}=\dfrac{4}{27}\cdot\dfrac{9}{2}=\dfrac{2}{3}

\bold{CORRECT}\\\\\dfrac{x}{\frac{4}{27}}=\dfrac{2}{3}\qquad\text{multiply both sides by}\ \dfrac{4}{27}\\\\x=\dfrac{2}{3}\cdot\dfrac{4}{27}\\\\x=\dfrac{8}{81}

8 0
3 years ago
Diane has 12 coins that total $1.60. If the coins are all nickels and
Dmitry_Shevchenko [17]

Answer:

she has 4 quarters because 4 quarters equals $1 and she has 12 nickles because 12 × 5 =60

6 0
2 years ago
Sums of a sequence: How many total gifts did my true love give to me during the entire 12 days of Christmas? Hint: Without any o
ale4655 [162]
<span> On the first day of Christmas,
my true love sent to me
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my true love sent to me
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The song continues, adding 4 calling birds on the 4th day, 5 golden rings on the 5th, and so on up to the 12th day, when 12 drummers add to the cacophony of assorted birds, pipers and lords leaping all over the place.

Notice that on each day there is one partridge (so I will have 12 partridges by the 12th day), and each day from the second day onwards there are 2 doves (so I will have 22 doves), and from the 3rd there are 3 hens (total of 30 hens), and so on.

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Total = 364

We observe that we have the same number of partridges as drummers (12 of each); doves and pipers (22 of each); hens and lords (30 of each) and so on. So the easiest way to count our presents is to add up to the middle of the list and then double the result: (12 + 22 + 30 + 36 + 40 + 42) × 2 = 364.


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3 years ago
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