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ipn [44]
3 years ago
6

You use 3 cans of soup to serve 5 people. If you want to serve 20 people, what should you do?

Mathematics
2 answers:
Elenna [48]3 years ago
6 0
<span>Most of the information’s that are required are already given in the question.</span>
The number of cans used to serve 5 people = 3
Then
The number of cans required to serve 1 people = (3/5) * 1
                                                                             = (3/5) cans
So
The number of cans required to serve 20 people = (3/5) * 20 cans
                                                                               = 3 * 4 cans
                                                                               = 12 cans
So 12 cans will be required to serve 20 people. I hope the procedure is clear enough for you to understand.



dusya [7]3 years ago
5 0
If you would like to know how many cans of soup you will need, you can calculate this using the following steps:

3 cans of soup ... 5 people
x cans of soup = ? ... 20 people

3 * 20 = 5 * x    /5
x = 3 * 20 / 5
x = 12 cans of soup

The correct result would be 12 cans of soup.
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The range of y=√x-5-1 is​
Lady_Fox [76]

Answer:

[-1, ∞ )

Step-by-step explanation:

The use of parentheses is essential here.  I am assuming that you meant:

y=√(x-5) - 1

Note that √(x-5) has the range [0, ∞ ).  The domain is [5, ∞ ).

Thus, y=√(x-5) - 1 has the range [-1, ∞ )

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3 years ago
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Simplify the expression.<br><br><br><br> 9y^2−25/9y^2−30y+25
mezya [45]

Answer:

Step-by-step explanation:

56y^2/9-30y+25

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3 years ago
-2/5 + 4/3 answer as a fraction
il63 [147K]

Answer:

your answer is 14/15.

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3 years ago
Co-ordinate Geometry
saul85 [17]

Answer:

The value of a  = 14

Step-by-step explanation:

Given

(x₁, y₁) = (2, 4)

(x₂, y₂) = (6, a)

(x₃, y₃) = (-1, 1)

A = 9 sq.units

Area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃) is:

A=\frac{\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|}{2}

substituting the values (x₁, y₁) = (2, 4), (x₂, y₂) = (6, a), (x₃, y₃) = (-1, 1), A = 9 in th formula

A=\frac{\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|}{2}

9=\frac{\left|2\left(a-1\right)+6\left(1-4\right)+-1\left(4-a\right)\right|}{2}

Multiply both sides by 2

\frac{2\left|2\left(a-1\right)+6\left(1-4\right)-1\left(4-a\right)\right|}{2}=9\cdot \:2

simplify

\left|2\left(a-1\right)+6\left(1-4\right)-1\left(4-a\right)\right|=18

As the area is always positive.

so

2\left(a-1\right)+6\left(1-4\right)-1\cdot \left(4-a\right)=18

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Add 18 to both sides

2\left(a-1\right)-18-\left(4-a\right)+18=18+18

simplify

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3a=42

Divide both sides by 3

\frac{3a}{3}=\frac{42}{3}

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7 0
3 years ago
Pls can any one solve this​
belka [17]

All the angles make a full circle which needs to equal 360 degrees.

Add all the angles together to equal 360:

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Combine like terms:

9x -30 = 360

Add 30 to both sides:

9x = 390

Divide both sides by 9:

X = 390/9

X = 43.3333. ( 43 and 1/3)

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