A square 20 cm long and a

Mathematics

1 answer:

Gemiola [76]2 years ago

4 0

Answer:

20 * 4 = 80

Step-by-step explanation:

the answer is 80 cuz you have to multiply 20 ( the square) by 4 ( the number of sides) and you get 80

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If I had 15 apples 18 oranges and 21 mangoes how many would go in each bag

Novosadov [1.4K]

Answer:

how many bags are there?

Step-by-step explanation:

6 0

3 years ago

Frank is using a cookie recipe that calls for 2 ¼ cups of flour. The recipe yields 20 cookies. How much flour would be needed to

larisa86 [58]

Hey there! I'm happy to help!

How many times does 20 go into 48?

48/20=2.4

So, we multiply our amount of flour for 20 cookies by 2.4.

2.25×2.4=5.4

Therefore you would need 5.4 or 5 2/5 cups of flour to make 48 cookies.

Have a wonderful day! :D

3 0

3 years ago

You are designing a rectangular garden. the garden will be enclosed by fencing on three sides and by a house on the fourth side.

RUDIKE [14]

Suppose the dimensions of the rectangle is x by y and let the side enclosed by a house be one of the sides measuring x, then the sides that is to be enclosed are two sides measuring y and one side measuring x.

Thus, the length of fencing needed is given by

P = x + 2y

The area of the rectangle is given by xy,

i.e.

$xy = 288 \\ \\ \\ \\ \Rightarrow y= \frac{288}{x}$

Substituting for y into the equation for the length of fencing needed, we have

$P=x+2\left( \frac{288}{x} \right)=x+ \frac{576}{x}$

For the amount of fencing to be minimum, then

$\frac{dP}{dx} =0 \\ \\ \Rightarrow1- \frac{576}{x^2} =0 \\ \\ \Rightarrow \frac{576}{x^2} =1 \\ \\ \Rightarrow x^2=576 \\ \\ \Rightarrow x=\sqrt{576}=24$

Now, recall that

$y= \frac{288}{x} = \frac{288}{24} =12$

Thus, the length of fencing needed is given by

P = x + 2y = 24 + 2(12) = 24 + 24 = 48.

Therefore, 48 feets of fencing is needed to enclose the garden.

8 0

3 years ago

Find and sketch the domain of the function. f(x,y)=√(4-x^2-y^2) +√(1-x^2)<br>

elena55 [62]

Answer:

Hello

Step-by-step explanation:

The domain is limited with 2 lines parallel: -1 ≤ x ≤ 1

and the disk ? (inside of a circle) of center (0,0) and radius 2

$dom\ f(x,y)=\{(x,y) \in \mathbb{R} ^2 | \ -1\leq x \leq -1\ and \ ( -\sqrt{4-x^2} \leq \ y \leq \sqrt{4-x^2}\ ) \ \}\\$