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3 years ago

URGENT EXPLANATION Solve the Inequality

Mathematics

1 answer:

SpyIntel [72]3 years ago

3 0

Answer:

3) a<5

5) z≤2

Step-by-step explanation:

3) a/5<1

a<1x5

a<5

5) 5(z+6)≤40

5z+30≤40

5z≤40-30

5z≤10

z≤10÷5

z≤2

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A point moves along the curve y = √ x in such a way that the y-component of the position of the point is increasing at a rate of

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Answer:

The value of x component changes at a rate of $\frac{dx}{dt}=4\sqrt{x}$ units per second

Step-by-step explanation:

We are given that $y=x^{\frac{1}{2}}$

Differentiating on both sides with respect to time we get

$\frac{dy}{dt}=\frac{d\sqrt{x}}{dt}\\\\\frac{dy}{dt}=\frac{1}{2}x^{\frac{-1}{2}}(\frac{dx}{dt})\\\\\frac{dy}{dt}=\frac{1}{2\sqrt{x}}\frac{dx}{dt}$

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Solving for $\frac{dx}{dt}$ we get

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2 years ago

Jen borrowed 1500 from her parents for car repair. She plans to take 10% of her $450 weekly income to pay them back. At this rat

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Answer:

10% of %450 is $45

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Step-by-step explanation:

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3 years ago

Mohamed decided to track the number of leaves on the tree in his backyard each year The first year there were 500 leaves Each ye

svetlana [45]

Answer:

The required recursive formula is

$f(n)= 500\times(1.4)^{n-1}\\$

Step-by-step explanation:

Mohamed decided to track the number of leaves on the tree in his backyard each year.

The first year there were 500 leaves

$Year \: 1 = 500$

Each year thereafter the number of leaves was 40% more than the year before so that means

$Year \: 2 = 500(1+0.40) = 500\times 1.4\\$

For the third year the number of leaves increase 40% than the year before so that means

$Year \: 3 = 500\times 1.4(1+0.40) = 500 \times 1.4^{2}\\$

Similarly for fourth year,

$Year \: 4 = 500\times 1.4^{2}(1+0.40) = 500\times 1.4^{3}\\$

So we can clearly see the pattern here

Let f(n) be the number of leaves on the tree in Mohameds back yard in the nth year since he started tracking it then general recursive formula is

$f(n)= 500\times(1.4)^{n-1}\\$

This is the required recursive formula to find the number of leaves for the nth year.

Bonus:

Lets find out the number of leaves in the 10th year,

$f(10)= 500\times(1.4)^{10-1}\\\\f(10)= 500\times(1.4)^{9}\\\\f(10)= 500\times20.66\\\\f(10)= 10330$

So there will be 10330 leaves in the 10th year.

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