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3 years ago

how many more times intense is an earthquake that measures 8 on the Ritcher scale when an earthquake that measures 5?

Mathematics

1 answer:

natta225 [31]3 years ago

6 0

Answer:

1000

Step-by-step explanation:

Since the Richter Scale is based off of an exponential equation, every point on the scale adds a magnitude of 10.

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An airplane flies 2700 miles in 3 hours and 45 minutes. What is its average speed in miles per hour?

fenix001 [56]

Answer:

720 MPH

Step-by-step explanation:

2700 miles / 3.75 hours = 720 MPH

6 0

3 years ago

Read 2 more answers

A solid oblique pyramid has a square base with edges measuring x cm. The height of the pyramid is (x + 2) cm. Which expression r

zavuch27 [327]

we know that

the volume of a solid oblique pyramid is equal to

$V=\frac{1}{3}*B*h$

where

B is the area of the base

h is the height of the pyramid

in this problem we have that

B is a square

$B=b^{2}$

where

$b=x\ cm$

$B=x^{2}\ cm^{2}$

$h=(x+2)\ cm$

substitute in the formula of volume

$V=\frac{1}{3}*x^{2}*(x+2)\\ \\V=\frac{1}{3}*[x^{3} +2x^{2}]\ cm^{3}$

therefore

the answer is

$V=\frac{1}{3}*[x^{3} +2x^{2}]\ cm^{3}$

4 0

3 years ago

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A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t

MakcuM [25]

Answer:

$165$ .

Step-by-step explanation:

Since repetition isn't allowed, there would be $11$ choices for the first donut, $(11 - 1) = 10$ choices for the second donut, and $(11 - 2) = 9$ choices for the third donut. If the order in which donuts are placed in the bag matters, there would be $11 \times 10 \times 9$ unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type $3 \times 2 \times 1 = 6$ times.

For example, if a bag includes donut of type $x$ , $y$ , and $z$ , the count $11 \times 10 \times 9$ would include the following $3 \times 2 \times 1$ arrangements:

$xyz$ .
$xzy$ .
$yxz$ .
$yzx$ .
$zxy$ .
$zyx$ .

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count $11 \times 10 \times 9$ by $3 \times 2 \times 1 = 6$ to find the actual number of donut combinations:

$\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of $3$ objects from a set of $11$ distinct objects:

$\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}$ .