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Aleksandr [31]
2 years ago
5

The table shows the amount of sugar needed to make lemonade. Are the quantities proportional? Explain which method you used and

how you used it to determine if they were proportionate or not.
Sugar (grams)

4

6

8

10

Servings

2

3

4

5
Mathematics
2 answers:
Klio2033 [76]2 years ago
7 0

Answer: Yes

Step-by-step explanation:

For the first sugar and servings part, 4 for sugar, 2 for servings.

The ratio for this example above it for sugars to servings it 2:1. So every 2 sugars is every 1 serving and this applies to every other of the sugars and servings in the table.

Mrrafil [7]2 years ago
6 0

Yes there are propotional.

If you pair them up in proportions:

4/2

6/3

8/4

10/5

If you divide them all you get 2 and that means they are all proportional

You might be interested in
It is given that A is directly proportional to r2 and r> 0. When r=5, A=75.
torisob [31]

Answer:

see explanation

Step-by-step explanation:

Given A is directly proportional to r² then the equation relating them is

A = kr² ← k is the constant of proportion

To find k use the condition when r = 5, A = 75 , then

75 = k × 5² = 25k ( divide both sides by 25 )

3 = k

A = 3r² ← equation of proportion

(a)

when r = 4, then

A = 3 × 4² = 3 × 16 = 48

(b)

when A = 147 , then

147 = 3r² ( divide both sides by 3 )

49 = r² ( take the square root of both sides )

r = \sqrt{49} = 7

3 0
3 years ago
Please help !! Picture provided !!!
adelina 88 [10]
Method One
Your calculator might be able to do this. Mine does it like this.
6
nCr
2  
= 
15

Method 2
You could simply set up 6C2

This gives you
\frac{n!}{(n- r)! r!} =  \frac{6!}{(6 - 2)!2!} =  \frac{6!}{4! 2!} =  \frac{6*5}{2*1} = 15

Method Three
You only have to do this a couple of times to see how the cancellation works.

\frac{6!}{4!2!} =  \frac{6*5*\st{4*3*2*1}}{4*3*2*1 * 2*1}
After all the cancellation takes place you have
6*5/2 = 15
7 0
3 years ago
What’s the value of x?
GaryK [48]

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7 0
3 years ago
A certain bridge arch is in the shape of half an ellipse 106 feet wide and 33.9 feet high. At what horizontal distance from the
nata0808 [166]

Answer:

The horizontal distance from the center is 49.3883 feet

Step-by-step explanation:

The equation of an ellipse is equal to:

\frac{x^2}{a^{2} } +\frac{y^2}{b^{2} } =1

Where a is the half of the wide, b is the high of the ellipse, x is the horizontal distance from the center and y is the height of the ellipse at that distance.

Then, replacing a by 106/2 and b by 33.9, we get:

\frac{x^2}{53^{2} } +\frac{y^2}{33.9^{2} } =1\\\frac{x^2}{2809} +\frac{y^2}{1149.21} =1

Therefore, the horizontal distances from the center of the arch where the height is equal to 12.3 feet is calculated replacing y by 12.3 and solving for x as:

\frac{x^2}{2809} +\frac{y^2}{1149.21} =1\\\frac{x^2}{2809} +\frac{12.3^2}{1149.21} =1\\\\\frac{x^2}{2809}=1-\frac{12.3^2}{1149.21}\\\\x^{2} =2809(0.8684)\\x=\sqrt{2809(0.8684)}\\x=49.3883

So, the horizontal distance from the center is 49.3883 feet

8 0
3 years ago
Please please please helpp
svp [43]

Answer:

I haven't done expressions in a while but I say the safest bet would be D.

6 0
2 years ago
Read 2 more answers
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