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3 years ago

I need help with question 7.

Mathematics

2 answers:

SIZIF [17.4K]3 years ago

8 0

Im really not sure but I think it’s A

Alika [10]3 years ago

8 0

I think it’s the first one

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Ally works in a shop for 18$ per hour. She is not allowed to work more than 40 hours a week but has to work atleast 20 hours. St

AleksAgata [21]

Answer:

Hi there!

The domain is:

20<=x<=40

Step-by-step explanation:

This states that:

x>=20

x<=40

Which fits the problem!

Hope this helps

8 0

3 years ago

X y

mario62 [17]

Let's prove each one:
A) y=x-3
x=0→y=0-3→y=-3=-3 ok
x=2→y=2-3→y=-1 different to 5 No

B) y=x+3
x=0→y=0+3→y=3 different to -3 No

C) y=4x-3
x=0→y=4(0)-3=0-3→y=-3=-3 ok
x=2→y=4(2)-3=8-3→y=5=5 ok
x=3→y=4(3)-3=12-3→y=9=9 ok
x=4→y=4(4)-3=16-3→y=13=13 ok

D) y=4x+3
x=0→y=4(0)+3=0+3→y=3 different to -3 No

Answer: Option C) y=4x-3

8 0

3 years ago

The population of a county is growing at a rate of 9% per year, compounded continuously. How many years will it take for the pop

torisob [31]

Answer:

$t=15.4\ years$

Step-by-step explanation:

The exponential growth function compounded continuously is equal to

$A=P(e)^{rt}$

where

A is the final population

P is the initial population

r is the rate of growth in decimal

t is Number of years

e is the mathematical constant number

we have

$A=4x\\P=x\\ r=9\%=9/100=0.09$

substitute in the function above

$4x=x(e)^{0.09t}$

simplify

$4=(e)^{0.09t}$

Take natural log of both sides

$ln(4)=ln[(e)^{0.09t}]$

$ln(4)=0.09t(ln(e))$

$ln(e)=1$

$ln(4)=0.09t$

$t=ln(4)/0.09$

$t=15.4\ years$

8 0

4 years ago

Tom invests 2 000dhs in a savings account where he earns 0.5% interest per year compounded

aleksklad [387]

analyze problem

complex interest

2000+2000*0.005=2010

2010+2010*0.005=2020.05

2020.05+2020.05*0.005=2030.15025

8 0

2 years ago

Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤

MrMuchimi

The average rate of change of a function f(x) in an interval, a < x < b is given by
$\frac{f(b) - f(a)}{b - a}$

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by $\frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4$

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by $\frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3$

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3$

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by $\frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1$

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by $\frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1$

6.) The average rate of change of q(x) in the interval -6 ≤ x ≤ 0 is given by $\frac{q(0)-q(-6)}{0-(-6)} = \frac{(0+3)^2-(-6+3)^2}{0+6} = \frac{9-9}{6} = \frac{0}{6} =0$

3 0

3 years ago

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