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3 years ago

The difference between 48 and 13 as a numérical or algebraic expression?

Mathematics

1 answer:

xenn [34]3 years ago

4 0

Its a numerical expression because it has no variables, no letters.

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What is the answer to 16> 1.5 + 0.8b

user100 [1]

Answer:

16>2.3b

Step-by-step explanation:

1.5+0.8=2.3

Substitute b as 1. Which would still keep it as 2.3.

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3 years ago

Please help I want summer school done. Which graph corresponds to the given function?

Alexus [3.1K]

Answer:

The first option in the first picture

Step-by-step explanation:

Good luck with summer school bud :)

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2 years ago

The number of customers that come to a certain clothing store each day follows a normal distribution. The mean number of custome

Lera25 [3.4K]

Answer:

c 2.5

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Suppose M is the midpoint

AysviL [449]

x = 9

Since M is the midpoint of XY:

XM = MY

XM = 2x + 11

MY = 5x - 16

2x + 11 = 5x - 16

5x - 2x = 11 + 16

3x = 27

x = 27/3

x = 8

Learn more here: brainly.com/question/18315903

6 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

3 years ago