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3 years ago

Finding the area of a figure

Mathematics

2 answers:

Novosadov [1.4K]3 years ago

7 0

The area is always the width x length :)

anzhelika [568]3 years ago

3 0

You can also use the formula Area=Length x Width or A=LxW

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Order the numbers from least to greatest. 0.3, 9/25, 33%, 0.35

Finger [1]

Answer:

0.3(least)

0.33

0.35

0.36(greatest)

Step-by-step explanation:

Okay, so first we need to have all of our numbers in the same category of variable. So first, we will convert 9/25 to a decimal.

To do that, we must divide 9 by 25 to get our answer of 0.36.

Now, we will convert the 33% to a decimal by dividing it by 100 to get 0.33

We now have our list, which should be ordered as:

0.3

0.33

0.35

0.36

5 0

2 years ago

Use benchmark fractions to compare 2/9 and 5/6. Part A Which of the following is a benchmark fraction Heather can use to compare

Dimas [21]

Answer:

Step-by-step explanation:

2/9 and 5/6 have a lcm of 30.

5 0

3 years ago

Read 2 more answers

What is the solution to this system of equations? -5.9x-3.7y=-2.1 and 5.9x+3.7y=2.1

goblinko [34]

No solution because everything cancels out

3 0

3 years ago

Select the basic integration formula you can use to find the indefinite integral.

kondaur [170]

Answer:

$\int \dfrac{du}{u}=\log|u|+C$

$\int u^n du=\dfrac{u^{n+1}}{n+1}+C$

$\int \dfrac{du}{u\sqrt{u^2-a^2}}=\dfrac{1}{a}\csc^{-1}(\dfrac{x}{a})+C$

Step-by-step explanation:

$\int \dfrac{du}{u}=\log|u|+C$ $[\because \int \dfrac{dx}{x}=\log |x|+C]$

$\int u^n du=\dfrac{u^{n+1}}{n+1}+C$ $[\because \int x^n dx=\dfrac{x^{n+1}}{n+1}+C]$

$\int \dfrac{du}{u\sqrt{u^2-a^2}}=\dfrac{1}{a}\csc^{-1}(\dfrac{x}{a})+C$ $[\because \dfrac{adx}{x\sqrt{x^2-a^2}}=\csc^{-1}(\dfrac{x}{a})+C]$

3 0

3 years ago

What does 2 tenths look like?

forsale [732]

Answer:

2/10 or 0.2

i dont know which tenths you wanted so i did both

7 0

3 years ago