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3 years ago

How to find the time

Mathematics

1 answer:

FrozenT [24]3 years ago

7 0

Use I=PRT/100 to solve the question

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Simplify.

Deffense [45]

Step-by-step explanation:

$( \frac{3}{2} ) ^{2} + 4 \div 2 - 3 \\ \frac{9}{4} + 4 \div 2 - 3 \\ \frac{9}{4} + 2 - 3 \\ \frac{9}{4} - 1 \\ \frac{9 - 4}{4} \\ \frac{5}{4}$

5 0

3 years ago

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Given the equation for the total surface area of a cylinder, solve for the height of the cylinder.

schepotkina [342]

H= (2pier^2)/ 2pier all subtracted by s so the last equation from the picture.
Hope it helps!

4 0

3 years ago

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Given f(x) = x² - 10x + 22, what is the range of f?

adoni [48]

Answer:

[-3, ∞)

Step-by-step explanation:

There are many ways to find the range but I will use the method I find the easiest.

First, find the derivative of the function.

f(x) = x² - 10x + 22

f'(x) = 2x - 10

Once you find the derivative, set the derivative equal to 0.

2x - 10 = 0

Solve for x.

2x = 10

x = 5

Great, you have the x value but we need the y value. To find it, plug the x value of 5 back into the original equation.

f(x) = x² - 10x + 22

f(5) = 5² - 10(5) + 22

= 25 - 50 +22

= -3

Since the function is that of a parabola, the value of x is the vertex and the y values continue going up to ∞.

This means the range is : [-3, ∞)

Another easy way is just graphing the function and then looking at the range. (I attached a graph of the function below).

Hope this helped!

6 0

3 years ago

Given parallelogram PARK.

Tanya [424]

Answer:

Step-by-step explanation:

5 0

3 years ago

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A rancher wants to fence in an area of 1500000 square feet in a rectangular field and then divide it in half with a fence down t

VLD [36.1K]

Answer:

6000 ft

Step-by-step explanation:

Let length of rectangular field=x

Breadth of rectangular field=y

Area of rectangular field= $1500000$ square ft

Area of rectangular field= $l\times b$

Area of rectangular field= $x\time y$

$1500000=xy$

$y=\frac{1500000}{x}$

Fencing used ,P(x)= $x+x+y+y+y=2x+3y$

Substitute the value of y

P(x)= $2x+3(\frac{1500000}{x})$

$P(x)=2x+\frac{4500000}{x}$

Differentiate w.r.t x

$P'(x)=2-\frac{4500000}{x^2}$

Using formula: $\frac{dx^n}{dx}=nx^{n-1}$

$P'(x)=0$

$2-\frac{4500000}{x^2}=0$

$\frac{4500000}{x^2}=2$

$x^2=\frac{4500000}{2}=2250000$

$x=\sqrt{2250000}=1500$

It is always positive because length is always positive.

Again differentiate w.r.t x

$P''(x)=\frac{9000000}{x^3}$

Substitute x=1500

$P''(1500)=\frac{9000000}{(1500)^3}>0$

Hence, fencing is minimum at x=1 500

Substitute x=1 500

$y=\frac{1500000}{1500}=1000$

Length of rectangular field=1500 ft

Breadth of rectangular field=1000 ft

Substitute the values

Shortest length of fence used= $2(1500)+3(1000)=6000 ft$

Hence, the shortest length of fence that the rancher can used=6000 ft

3 0

4 years ago