2 years ago

Please answer to this question

Mathematics

1 answer:

Marta_Voda [28]2 years ago

5 0

Answer:

Step-by-step explanation:

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What is the answer to this equatiion? ln*20+ln*5=2*ln*x

Rudiy27

Ln20+ln5=2lnx
ln(20*5)=lnx^2
ln100=lnx^2
100=x^2
x=+10 or -10

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3 years ago

Which of the following is equivalent to 10000 lb?

dalvyx [7]

Answer:

5 Tons

Step-by-step explanation:

Or 7,858.35 Pound sterling In Money Standards

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2 years ago

Suppose you have a cube of length s. The volume of that cube is V = s^3. Now let's suppose the dimensions of that cube (and henc

Grace [21]

Answer:

Part 1

s'(t)=1

$V'(t)=3t^2$

Part 2

$s'(t)=\frac{1}{3\sqrt[3]{t^2}}$

V'(t)=1

Step-by-step explanation:

As stated in the question, if s is the length of a cube, its volume is

$V(s)=s^3$

When s changes in time, the volume will change too. The challenge here is to find the growth of the volume in terms of those changes in length

Part 1

$s(t)=t$

The change of s with respect to time is found by differentiating the relation to get

$s'=1$

The volume will also change, and its derivative is

$V'(s)=3s^2.s'$

Since s'=1

$V'(s)=3s^2$

How s=t

$V'(t)=3t^2$

Part 2

$V(t)=t$

Replacing this into the formula for V(s)

$V(s)=s^3=t$

So we have

$s=\sqrt[3]{t}=t^{\frac{1}{3}}$

Computing the derivatives:

V(t)=t =>

V'(t)=1

$s'(t)=\frac{1}{3}t^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{t^2}}$

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3 years ago

Suppose that $7000 is invested at an interest rate of 5.2% per year, compounded

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Answer:

use the formula 14000=7000(1+5.2÷100)^t-1

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2 years ago

Q. 2 Find the HCF of 16, 44 and 84 by listing the factors.

ollegr [7]

The HCF is 4. Use ladder method to find it.

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1 year ago

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