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3 years ago

This one also I needed help on

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

5 0

Answer:

y = x³

Step-by-step explanation:

Note that

1³ = 1 × 1 × 1 = 1 ← value of y- coordinate

2³ = 2 × 2 × 2 = 4 × 2 = 8 ← value of y- coordinate

3³ = 3 × 3 × 3 = 9 × 3 = 27 ← value of y- coordinate

4³ = 4 × 4 × 4 = 16 × 4 = 64 ← value of y- coordinate

5³ = 5 × 5 × 5 = 25 × 5 = 125 ← value of y- coordinate

The rule is that the y- coordinate is the cube of the x- coordinate, that is

y = x³

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30 pts ! The height of the Ferris wheel in feet can be represented by the function 50 cos (π/15(x-10))+52 with time in x seconds

Dahasolnce [82]

h = 50 cos ( pie(x - 10 )/15 ) + 52

80 = 50 cos ( pie( x - 10 )/15 ) + 52

80 - 52 = 50 cos ( pie( x - 10 )/15 )

28 = 50 cos ( pie( x - 10 )/15 )

cos ( pie( x - 10 )/15 ) = 28/50

cos ( pie( x - 10 )/15 ) = 56/100

cos ( pie( x - 10 )/15 ) = cos ( 56 )

cos ( pie( x - 10 )/15 ) = cos ( 0.3111 pie )

Thus ;

pie( x - 10 )/15 = 0.3111 pie

( x - 10 )/15 = 0.3111

x - 10 = 15 × 0.3111

x - 10 = 4.6665

x = 10 + 4.6665

x = 14.6665 [ approximately ]

Thus the correct answer is exactly what u chose goodjob .....

7 0

3 years ago

The population, P(t), of China, in billions, can be approximated by1 P(t)=1.394(1.006)t, where t is the number of years since th

vitfil [10]

Answer:

At the start of 2014, the population was growing at 8.34 million people per year.

At the start of 2015, the population was growing at 8.39 million people per year.

Step-by-step explanation:

To find how fast was the population growing at the start of 2014 and at the start of 2015 we need to take the derivative of the function with respect to t.

The derivative shows by how much the function (the population, in this case) is changing when the variable you're deriving with respect to (time) increases one unit (one year).

We know that the population, P(t), of China, in billions, can be approximated by $P(t)=1.394(1.006)^t$

To find the derivative you need to:

$\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=\\\\\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'\\\\1.394\frac{d}{dt}\left(1.006^t\right)\\\\\mathrm{Apply\:the\:derivative\:exponent\:rule}:\quad \frac{d}{dx}\left(a^x\right)=a^x\ln \left(a\right)\\\\1.394\cdot \:1.006^t\ln \left(1.006\right)\\\\\frac{d}{dt}\left(1.394\cdot \:1.006^t\right)=(1.394\cdot \ln \left(1.006\right))\cdot 1.006^t$

To find the population growing at the start of 2014 we say t = 0

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(0)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^0\\P(0)' = 0.00833901 \:Billion/year$

To find the population growing at the start of 2015 we say t = 1

$P(t)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^t\\P(1)' = (1.394\cdot \ln \left(1.006\right))\cdot 1.006^1\\P(1)' = 0.00838904 \:Billion/year$