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Igoryamba
3 years ago
9

1. If a cars speed increases from 18 m/s to 20 m/s in 15 seconds, its acceleration is how much :

Mathematics
1 answer:
Paraphin [41]3 years ago
8 0

Answer: Q1: 0.13m/s^2.

Q2: 1.2m/s^2

Q3: C

Q4: 2.8m/s

Q5: 0.63m/s

Q6: 2040km

Step-by-step explanation: acceleration a = (v-u)/t

Q1: a =(20-18)/15

a= 0.13m/s^2

Q2: a =(0.12-0)/0.1

a= 1.2m/s^2

Q3: None

Q4: V= distance/time

V= 12.6/4.5

V= 2.8m/s

Q5: v=2.5/4

v= 0.63m/s

Q6: distance d = v*t

d =85*24

d= 2040km

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<u>Given </u>that the functions f(x)=x+4 and g(x)=x^{3}

We need to determine the value of the function (g \ {\circ} f)(-3)

First, we shall determine the composition of the function (g \circ f)(x)

<u>Function </u>(g \circ f)(x)<u>:</u>

Let us determine the function (g \circ f)(x)

Thus, we have;

(g \circ f)(x)=g[f(x)]

               =g[x+4]

               =(x+4)^3

(g \circ f)(x)=x^3+3x^2(4)+3x(4)^2+(4)^3

(g \circ f)(x)=x^3+12x^2+48x+64

Thus, the function is (g \circ f)(x)=x^3+12x^2+48x+64

<u>Value of the function </u>(g \ {\circ} f)(-3)<u>:</u>

The value of the function can be determined by substituting x = -3 in the function (g \circ f)(x)=x^3+12x^2+48x+64

Thus, we have;

(g \circ f)(-3)=(-3)^3+12(-3)^2+48(-3)+64

Simplifying the terms, we get;

(g \circ f)(-3)=-27+12(9)+48(-3)+64

(g \circ f)(-3)=-27+108-144+64

(g \circ f)(-3)=1

Thus, the value of the function (g \ {\circ} f)(-3) is 1.

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