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2 years ago

What is the mass of 20 buckets

Mathematics

2 answers:

Margarita [4]2 years ago

5 0

10 percent is the mass number

Olenka [21]2 years ago

4 0

20xmass of one bucket

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A circle is placed in a square with a side length of 18 mm, as shown below. Find the area of the shaded region.

finlep [7]

Answer: the shaded region = 69.66mm^2

Explanation:
Assume shaded region is X
Then :
X + Area of circle = area of square
X = area of square - area of circle

Find area of square:
Side length = 18mm
Area = 18 x 18 = 324

Find area of circle:
Side length of a square = the diameter of the circle = 18mm
And r(radius) = d/2 = 18/2 = 9
Area = pi x 9^2 = 254.34

Now find X
X = 324 - 254.34 = 69.66

3 0

2 years ago

Work out area of this semi circle with diameter 10

ANTONII [103]

Area of a circle is pi x r^2

R is the radius which is have the diameter: r = 10/2 = 5

Area of the full circle would be 3.14 x 5^2 = 3.14 x 25 = 78.5 square units.

A semi circle is half a full circle

Area of the semi circle = 78.5/2 = 39.25 square units.

8 0

3 years ago

Please please please help me

alexira [117]

A and can I be brainliest? :3

7 0

3 years ago

Suppose f(x) = 6x^2 + 2x − 7 and g(x) = 4x − 3. Find each of the following functions.

FinnZ [79.3K]

Step-by-step explanation:

(f+g)(x) ➡ f(x) + g(x) ➡ 6x^2 + 2x -7 + 4x - 3 =6x^2 + 6x - 10

(f-g)(x) ➡ f(x) - g(x) ➡ 6x^2 + 2x - 7 -4x + 3 = 6x^2 -2x -4

8 0

3 years ago

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BabaBlast [244]

$\underline{\bf{Given \:equation:-}}$

$\\ \sf{:}\dashrightarrow ax^2+by+c=0$

$\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.$

$\sf We\:know,$

$\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}$

$\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}$

$\underline{\large{\bf Identities\:used:-}}$

$\boxed{\sf (a+b)^2=a^2+2ab+b^2}$

$\boxed{\sf (√a)^2=a}$

$\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}$

$\boxed{\sf \sqrt{\sqrt{a}}=a}$

$\underline{\bf Final\: Solution:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}$

$\bull\sf Apply\: Squares$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2= (\sqrt{\alpha})^2+2\sqrt{\alpha}\sqrt{\beta}+(\sqrt{\beta})^2$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}$

$\bull\sf Put\:values$

$\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}$

$\bull\sf Simplify$

$\\ \sf{:}\dashrightarrow \underline{\boxed{\bf {\sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\sqrt{\dfrac{-b}{a}}+\sqrt{2}\dfrac{c}{a}}}}$

$\underline{\bf More\: simplification:-}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}$

$\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}$

$\underline{\Large{\bf Simplified\: Answer:-}}$

$\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}$

5 0

2 years ago

Read 2 more answers