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Ghella [55]
3 years ago
8

What is the domain and range function?

Mathematics
1 answer:
VLD [36.1K]3 years ago
3 0

Answer:

A.

Step-by-step explanation:

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When a constant force acts upon an object, the acceleration of the object varies inversely with its mass. When a certain constan
vitfil [10]

Answer:

a = 6m/s^2

Step-by-step explanation:

Given

When mass = 4kg; Acceleration = 15m/s²

Required

Determine the acceleration when mass = 10kg, provided force is constant;

Represent mass with m and acceleration with a

The question says there's an inverse variation between acceleration and mass; This is represented as thus;

a\ \alpha\ \frac{1}{m}

Convert variation to equality

a = \frac{F}{m}; Where F is the constant of variation (Force)

Make F the subject of formula;

F = ma

When mass = 4kg; Acceleration = 15m/s²

F = 4 * 15

F = 60N

When mass = 10kg; Substitute 60 for Force

F = ma

60 = 10 * a

60 = 10a

Divide both sides by 10

\frac{60}{10} = \frac{10a}{10}

a = 6m/s^2

<em>Hence, the acceleration is </em>a = 6m/s^2<em />

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3 years ago
Rewrite 9.524 as a mixed number in lowest terms
Anit [1.1K]

Answer:   9 132

                 ----------

                    250

9.524 = 9 + 0.524 = 9 + 524       =  9 524 =    9 524:2 = 9 262 =   9 134

                                      -----------       -----------       -----------    ---------     ---------

                                        1000             1000        1000:2     500          250

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Answer:

(-\infty,0)\cup(\frac{16}{3},\infty)\\That \ is x\frac{16}{3}

Step-by-step explanation:

h(x)=\frac{1}{8}x^{3} -x^{2} \\\\Differentiate \ h(x) \ with \  respect \ to \ x\\h'(x)=\frac{3}{8}x^{2} -2x

For positive rate of change h'(x)>0

\frac{3}{8} x^{2} -2x>0\\x(\frac{3}{8}x-2)>0\\\\ When \ x0 \ (multiplication \ of \ two \ positives \ is \ positive)\\\\h'(x)>0 \ when \ x\frac{16}{3} \\\\

6 0
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