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3 years ago

What is the slope of the line that passes through the points (10, 8) and (-15, 18)?

Mathematics

1 answer:

nataly862011 [7]3 years ago

6 0

Answer:

-2/5

Step-by-step explanation:

We can use the slope formula

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

= ( 18-8)/( -15-10)

= 10/-25

-2/5

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Find the value of expression m(m-n) for m=3 and n=1 I need the ANSWER ASAP

Alex73 [517]

Answer:

Step-by-step explanation:

m = 3

n = 1

Equation

m(m - n) Substitute the givens

Solution

3(3 -1 ) Evaluate what is inside the brackets.

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3 years ago

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Katarina [22]

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Step-by-step explanation:

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2 years ago

Plz help asappppppppp!!!

MAVERICK [17]

Answer:

10, 22, 48.4, 106.48, 234.256

Step-by-step explanation:

The first term is already given it is 10.

The second term is the first term times 2.2.

10×2.2=22

The second term is 22.

The third term is the second term times 2.2.

22×2.2=48.4

The third term is 48.4.

The fourth term is the third term times 2.2.

48.4×2.2=106.48

The fourth term is 106.48.

The fifth term is the fourth term times 2.2.

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The fifth term is 234.256.

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3 years ago

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AleksandrR [38]

Answer:

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Step-by-step explanation:

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2 years ago

A leprechaun places a magic penny under a girls pillow. The next night there are 2 magic pennies under her pillow. The following

kiruha [24]

Answer:

After 47 days she will have more than 90 trillion pennies.

Step-by-step explanation:

At the beginning there was 1 penny. At the second day the amount of pennies under the pillow became 2.

The amount of pennies doubled each day. So the series is,

$1,2,4,8,16,32,.....$

This series is in geometric progression.

As the pennies from each of the previous days are not being stored away until more pennies magically appear so the sum of series will be,

$S_n=\dfrac{a(r^n-1)}{r-1}$

where,

a = initial term = 1,

r = common ratio = 2,

As we have find the number of days that would elapse before she has a total of more than 90 trillion, so

$\Rightarrow 90\times 10^{12}\le \dfrac{1(2^n-1)}{2-1}$

$\Rightarrow 90\times 10^{12}\le \dfrac{2^n-1}{1}$

$\Rightarrow 90\times 10^{12}\le 2^n-1$

$\Rightarrow 2^n\ge 90\times 10^{12}+1$

$\Rightarrow \log 2^n\ge \log (90\times 10^{12}+1)$

$\Rightarrow n\times \log 2\ge \log (90\times 10^{12}+1)$

$\Rightarrow n \ge \dfrac{\log (90\times 10^{12}+1)}{\log 2}$

$\Rightarrow n \ge 46.4$

$\Rightarrow n\approx 47$

8 0

3 years ago