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2 years ago

Which pair shows equivalent expressions?

Mathematics

1 answer:

777dan777 [17]2 years ago

5 0

Answer:

3(x + 2) = 3x + 6

Step-by-step explanation:

3(x + 2) = 3x + 6

Distribute --> 3x + 6 = 3x + 6

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I need help please i'll give brainly

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Answer:

60 +24 = 84

Step-by-step explanation:jjjjjhjhjhj

6 x 4 +5 x 12

7 0

3 years ago

Caitlin has 4 and 3/4 lb of place use 1 1/10 pound to make a cup and another 2 pounds to make a jar how many pounds are left

g100num [7]

We know that
4 3/4---------> (4*4+3)/4----------> 19/4
1 1/10-----------> (10*1+1)/10----> 11/10

[pounds left]=[(19/4)-(11/10)-(2)]----> (10*19-4*11-40*2)/40
=(190-44-80)/40-------> 66/40---------> 1.65 pounds
1.65-----> 1 65/100--------> 1 13/20 pounds

the answer is 1.65 pounds (1 13/20 pounds)

7 0

3 years ago

The perimeter of a rectangle is 50 cm and the length is 4 cm longer than the width. Write and solve the system of equations usin

k0ka [10]

P=2(l+w)
50=2(4+w)
50=8+2w
collect the like terms 50-8=2w
42=2w divide both sides by 2
w=21

6 0

3 years ago

densk [106]

Camila's data set has the greatest interquartile range and the value of IQR = 8 option third is correct.

<h3>What is the range?</h3>

It is defined as the difference between the maximum value in the data set to the minimum value in the data set.

We have:

The amount of money in tips earned by four restaurant servers waiting on 10 tables is represented by the following data sets:

Alyssa {3, 6, 2, 8, 12, 14, 5, 7, 7, 8}

Bryant {9, 2, 7, 50, 0, 5, 2, 8, 6, 8}

Camila {1, 9, 10, 3, 0, 12, 10, 9, 8, 2}

Devon {4, 2, 8, 15, 20, 7, 5, 0, 6, 2}

The IQR for Alyssa:

IQR = Q3 - Q1

IQR = 8 - 5 = 3

The IQR for Bryant:

IQR = Q3 - Q1

IQR = 8 - 2 = 6

The IQR for Camila:

IQR = Q3 - Q1

IQR = 10 - 2 = 8

The IQR for Devon:

IQR = Q3 - Q1

IQR = 8 - 2 = 6

Thus, Camila's data set has the greatest interquartile range and the value of IQR = 8 option third is correct.

Learn more about the range here:

brainly.com/question/17553524

#SPJ1

8 0

2 years ago

Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1

aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

${3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}$

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

$3y^{2}.\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x}{y^{5}}$

3 0

2 years ago