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3 years ago

Which of the following expressions is equivalent to 4(m – 16)?

Mathematics

1 answer:

Masteriza [31]3 years ago

3 0

Answer:

4m - 64

Step-by-step explanation:

Solve 4(m - 16)

4(m - 16) = 4 x m - 16 x 4

= 4m - 64

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What is 0.1554 divided by 74?

umka2103 [35]

Answer:

0.0021

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

8.45 as a mixed number

Vlad [161]

A mixed number is a combination of a whole number and a fraction, so in this case we have 8 as the whole part and 0.45 as the fraction, 0.45 is equal to 45/100, and that fraction can be simplified as well, lets see:
8.45 = 8 45/100
= 8 9/20

4 0

2 years ago

There are values of t so that sin t=.35 and cos t=.6

NISA [10]

Recall the fundamental rule of trig:

$\sin^2(x)+\cos^2(x)=1 \quad\forall x \in \mathbb{R}$

So, there exists an angle $t$ such that

$(0.6,0.35)=(\sin(t),\cos(t))$

if and only if

$\sin^2(t)+\cos^2(t)=0.6^2+0.35^2=1$

Working out the numbers, we get

$0.6^2+0.35^2=0.36+0.1225=0.4825\neq 1$

So, there doesn't exist a number $t$ such that

$(0.6,0.35)=(\sin(t),\cos(t))$

7 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:

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

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