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2 years ago

Write an equation of the line that passes through the points (3,6) and (4,9)

Mathematics

1 answer:

ololo11 [35]2 years ago

3 0

Answer:

y=3x-3

Step-by-step explanation:

slope = change in y/change in x

slope = 3/1 = 3

y-intercept = y value when x=0

go 3u left, 3x3 = 9 so 9u down

(0,-3)

slope = m = 3

y-int = b = -3

y=mx+b

y=3x-3

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(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

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2 years ago

Which shows the list of numbers in order from least to greatest?

nirvana33 [79]

Answer:

-1, |-2|, |-3|, |3.5|, |3²|

Step-by-step explanation:

Calculate the absolute value (the absolute value of any number is always positive): 3, 3², 2 , 3.5, -1

Calculate the exponents : 3, 9, 2 , 3.5, -1

Put in order least to greatest : -1, |-2|, |-3|, |3.5|, |3²|

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2 years ago

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SVETLANKA909090 [29]

Step-by-step explanation:

then poured into another container with a rectangular base of length 27 cm and width 11 cm. Calculate the depth (height) of the water in the rectangular container

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2 years ago

What's the algebraic expression for this phrase: <br> the quotient of 100 and the quantity 6 plus w

leonid [27]

100/6 + w would be correct answer I think.

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3 years ago

Which of the following expressions is equal to 5 to the 6th power divided by 5 to the 5th power

Jobisdone [24]

Asey

remember
$\frac{x^m}{x^n}=x^{m-n}$
so

$\frac{5^6}{5^5}=5^{6-5}=5^1=5$

8 0

2 years ago

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