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Irina18 [472]
4 years ago
14

How do I solve this 8(c-9)=6(2c-12)-4c

Mathematics
1 answer:
bonufazy [111]4 years ago
6 0
First you are going to distribute and you are going to get 8c-72=12c-72-4c. Next you are going to combine like terms you get 8c-72=8c-72. Then move the Varible to one side and solve for the Varible you should get c=0
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A 2-column table with 5 rows. Column 1 is labeled Number of chores, x with entries 2, 4, 6, 8, 10. Column 2 is labeled Days, y w
charle [14.2K]

The constant of variation based on the information is one half.

<h3>How to illustrate the information?</h3>

From the information given, when x is 2, y is 1, when x is 4, y is 2, etc.

Therefore, the constant of variation will be:

= y/x

= 1/2 = 2/4

= 1/2 = 1/2

Therefore, the constant of variation based on the information is one half.

Learn more about constant of variation on:

brainly.com/question/25215474

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3 0
2 years ago
Another way of writing 6^2
kolezko [41]
Another way of writing 6 to the 2nd power would be, B, 6*6 . all it is is just multiplying by it's self however many times the power says to . <span />
8 0
4 years ago
Read 2 more answers
Which ordered pairs are in the solution set of the system of linear inequalities?
Marianna [84]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Please help me with question number 2
Mice21 [21]

Answer:

2. The area of the side walk is approximately 217 m²

3. The distance away from the sprinkler the water can spread is approximately 11 feet

4. The area of the rug is 49.6

Step-by-step explanation:

2. The dimensions of the flower bed and the sidewalks are;

The diameter of the flower bed = 20 meters

The width of the circular side walk, x = 3 meters

Therefore, the diameter of the outer edge of the side walk, D, is given as follows

D = d + 2·x (The width of the side walk is applied to both side of the circular diameter)

∴ D  = 20 + 2×3 = 26

The area of the side walk = The area of the sidewalk and the side walk = The area of the flower bed

∴ The area of the side walk, A = π·D²/4 - π·d²/4

∴ A = 3.14 × 26²/4 - 3.14 × 20²/4 = 216.66

By rounding to the nearest whole number, the area of the side walk, A ≈ 217 m²

3. Given that the area formed by the circular pattern, A = 379.94 ft.², we have;

Area of a circle = π·r²

∴ Where 'r' represents how far it can spread, we have;

π·r² = 379.94

r = √(379.94 ft.²/π) ≈ 10.997211 ft.

Therefore, the distance away from the sprinkler the water can spread, r ≈ 11 feet

4. The circumference of the rug = 24.8 meters

The circumference of a circle, C = 2·π·r

Where;

r = The radius of the circle

π = 3.1

∴ For the rug of radius 'r', C = 2·π·r = 24.8

r = 24.8/(2·π) = 12.4/π = 12.4/3.1 = 4

The area = π·r²

∴ The area of the rug = 3.1 × 4² = 49.6.

8 0
3 years ago
(w^2 – 6w)^2– 2(w^2 – 6w) – 35 = 0​
muminat

Step-by-step explanation:

( {w}^{2}  - 6w {)}^{2}  - 2( {w}^{2}  - 6w) - 35 = 0 \\  =  {w}^{4} - 6 {w}^{2}   - 2 {w}^{2}  - 12w  =  35  \\  =  {w}^{4}   - 4 {w}^{2}  - 12w = 35

4 0
4 years ago
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