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3 years ago

Write the following expression using a single exponent. 85 ÷ 8-2 A. 8-7 B. 83 C. 8-10 D. 87

Mathematics

1 answer:

attashe74 [19]3 years ago

7 0

The answer is A. I'm think i'm sure. :)

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Why math even exist people could just like wing it for everything

s2008m [1.1K]

Answer:

- 7 / 4

Step-by-step explanation:

( x1, y1 ) = ( 1, 10 )

Here,

x1 = 1

y1 = 10

( x2, y2 ) = ( 5, 3 )

Here,

x2 = 5

y2 = 3

Formula : -

Slope = ( y2 - y1 ) / ( x2 - x1 )

Slope = ( 3 - 10 ) / ( 5 - 1 )

Slope = - 7 / 4

Proper fraction = - 7 / 4

8 0

3 years ago

Read 2 more answers

You need 10,000 to pay for your final college tuition. you find a loan it is compounded quartely at 6.25% for 10 years. how much

Amanda [17]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$10000\\
r=rate\to 6.25\%\to \frac{6.25}{100}\to &0.0625\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &10
\end{cases}
\\\\\\
A=10000\left(1+\frac{0.0625}{4}\right)^{4\cdot 10}
\\\\\\
A=10000(1.015625)^{40}\implies A\approx 18592.39366677226890$

3 0

3 years ago

Find the area of rectangle ABCD

satela [25.4K]

9514 1404 393

Answer:

D. 12

Step-by-step explanation:

There are a number of ways to find the area of this rectangle. Perhaps the most straightforward is to find the lengths of the sides and multiply those. The distance formula is useful.

d = √((x2 -x1)^2 +(y2 -y1)^2)

Using the two upper-left points, we find the length of that side to be ...

d = √((3 -0)^2 +(3 -0)^2) = √(9 +9) = √18 = 3√2

Similarly, the length of the lower-left side is ...

d = √((-2 -0)^2 +(-2 -0)^2) = √(4+4) = √8 = 2√2

Then the area of the rectangle is ...

A = LW

A = (3√2)(2√2) = 3·2·(√2)^2 = 3·2·2 = 12

The area of rectangle ABCD is 12.

_____

Other methods include subtracting the area of the corner triangles from the area of the bounding square:

5^2 -2(1/2)(3·3) -2(1/2)(2·2) = 25 -9 -4 = 12

4 0

3 years ago

What is the sum of 2/11’ 2/7’ and 1/11’?

pav-90 [236]

Solution

To solve this addition of fractions we are gonna apply the fraction rule which is;

a/c + b/c = (a + b)/c

2/11 + 1/11 = (2 + 1)/11

= (2 + 1)/11 + 2/7

Now add the numerators 2 + 1

= (2 + 1)/11 + 2/7

= 3/11 + 2/7

Now we find the least common multiple

The least common multiple of 11 and 7 is 77.

But before that we cross multiply the numerators.

3 * 7 = 21
2 * 11 = 22

Now the ajustes fraction is;

21/77 + 22/77

Apply the fraction rule which is;

a/c + b/c = (a + b)/c

= (21 + 22)/77

Now we add the numerators.

43/77

Therefore the answer in fraction is 43/77 and in decimal form it is 0.55841

6 0

3 years ago

In a population of 10,000, there are 5000 nonsmokers, 2500 smokers of one pack or less per day, and 2500 smokers of more than on

Kazeer [188]

Answer:

In one month, we will have 4,950 non-smokers, 2,650 smokers of one pack and 2,400 smokers of more than one pack.

In two months, we will have 4,912 non-smokers, 2,756 smokers of one pack and 2,332 smokers of more than one pack.

In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

Step-by-step explanation:

We have to write the transition matrix M for the population.

We have three states (nonsmokers, smokers of one pack and smokers of more than one pack), so we will have a 3x3 transition matrix.

We can write the transition matrix, in which the rows are the actual state and the columns are the future state.

- There is an 8% probability that a nonsmoker will begin smoking a pack or less per day, and a 2% probability that a nonsmoker will begin smoking more than a pack per day. Then, the probability of staying in the same state is 90%.

- For smokers who smoke a pack or less per day, there is a 10% probability of quitting and a 10% probability of increasing to more than a pack per day. Then, the probability of staying in the same state is 80%.

- For smokers who smoke more than a pack per day, there is an 8% probability of quitting and a 10% probability of dropping to a pack or less per day. Then, the probability of staying in the same state is 82%.

The transition matrix becomes:

$\begin{vmatrix} &NS&P1&PM\\NS& 0.90&0.08&0.02 \\ P1&0.10&0.80 &0.10 \\ PM& 0.08 &0.10&0.82 \end{vmatrix}$

The actual state matrix is

$\left[\begin{array}{ccc}5,000&2,500&2,500\end{array}\right]$

We can calculate the next month state by multupling the actual state matrix and the transition matrix:

$\left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4950&2650&2400\end{array}\right]$

In one month, we will have 4,950 non-smokers, 2,650 smokers of one pack and 2,400 smokers of more than one pack.

To calculate the the state for the second month, we us the state of the first of the month and multiply it one time by the transition matrix:

$\left[\begin{array}{ccc}4950&2650&2400\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] =\left[\begin{array}{ccc}4912&2756&2332\end{array}\right]$

In two months, we will have 4,912 non-smokers, 2,756 smokers of one pack and 2,332 smokers of more than one pack.

If we repeat this multiplication 12 times from the actual state (or 10 times from the two-months state), we will get the state a year from now:

$\left( \left[\begin{array}{ccc}5000&2500&2500\end{array}\right] * \left[\begin{array}{ccc}0.90&0.08&0.02\\0.10&0.80 &0.10\\0.08 &0.10&0.82\end{array}\right] \right)^{12} =\left[\begin{array}{ccc}4792.63&3005.44&2201.93\end{array}\right]$

In a year, we will have 4,793 non-smokers, 3,005 smokers of one pack and 2,202 smokers of more than one pack.

3 0

4 years ago