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3 years ago

Order these fractions from least to greatest 3/8, 2/3, 4/5

Mathematics

1 answer:

Ivan3 years ago

6 0

Make all the denominators equal
3/8 = 45/120
2/3 = 80/120
4/5 = 96/120
From least to greatest we have,
3/8, 2/3, 4/5

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I just need help with this one question

Arte-miy333 [17]

Answer: 1/4 chance of picking red

Step-by-step explanation: There are 120 buttons in all, so the fraction of red marbles is 30/120. 30/120 = 1/4

Hope this helps!

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3 years ago

Read 2 more answers

Find the coordinates of the midpoint of the segment whose endpoints are R(9, 3) and S(-1, -9).

amm1812

Answer:

$(4, -3 )$

General Formulas and Concepts:

Pre-Algebra I

Order of Operations: BPEMDAS

Algebra I

Midpoint Formula: $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} )$

Step-by-step explanation:

Step 1: Define

R (9, 3)

S (-1, -9)

Step 2: Find midpoint

Substitute: $(\frac{9-1}{2}, \frac{3-9}{2} )$
Subtract: $(\frac{8}{2}, \frac{-6}{2} )$
Divide: $(4, -3 )$

4 0

3 years ago

Evaluate 6 P 3. 720 240 540 120

Damm [24]

Answer:120

Step-by-step explanation: i know i am a week late but ...

3 0

4 years ago

Given h(x) = 3x + 4, find h(-1).

larisa86 [58]

Plug in -1 for x
h(-1) = 3(-1) + 4
h(-1) = -3 + 4
Solution: h(-1) = 1

6 0

3 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

3 years ago