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Andrew [12]
3 years ago
14

Solve for x. 3(3x - 1) + 2(3 - x) = 0

Mathematics
2 answers:
olganol [36]3 years ago
7 0
Use distributive property to get rid of the parenthesis.....

9x - 3 + 6 - 2x = 0
combine like terms
7x + 3 = 0
move the 3 over by subtraction
7x = -3
divide by 7
x = -3/7
Snowcat [4.5K]3 years ago
4 0
3×3 is 9/ 3×-1 is-3
2×3 is 6/ 2×-1 is-2
9x-3+6-2x=0

Then add common components
7x+3=0
-3 on each side
7x=-3
Divide by 7 on each side
X=-0.428
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Solve for x in the equation x squared + 10 x + 12 = 36.
Anna35 [415]
It’s the first one

x = -12 or x=2
6 0
3 years ago
How to solve this.im stuck in this question for 20 mins. Plzz answer
Alik [6]

Answer:

Find the minimum or maximum value of the function g (I) = -3x^2 - 6x + 5. Describe the domain and range of the function, and where the function is increasing and decreasing. > -1 all real numbers The function The maximum value is I < 0 The domain is and the range is right of I left -1 is increasing to the of I= and decreasing to the 0 12 :: yo y0 :: 8 :: IS-1 :: -1 :: 0 :: I> 0 :: 1 :: 8 :: < 0 :: left :: all real numbers :: y -8 :: y < 8 :: y8​

Step-by-step explanation:

4 0
2 years ago
25% of what number is 15. Can u please explain how you solved it. Thanks :)
Ilya [14]
60.
let x be the number you're trying to solve.
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4 0
3 years ago
Approximate the change in the volume of a sphere when its radius changes from r​ = 40 ft to r equals 40.05 ft (Upper V (r )equal
alexgriva [62]

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere (V), measured in cubic feet, is represented by the following formula:

V = \frac{4\pi}{3}\cdot r^{3}

Where r is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

\Delta V = \frac{\partial V}{\partial r}\Delta r

The derivative of the volume as a function of radius is:

\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}

Then, the change in volume is expanded:

\Delta V = 4\pi \cdot r^{2}\cdot \Delta r

If r = 40\,ft and \Delta r = 40\,ft-40.05\,ft = 0.05\,ft, the change in the volume of the sphere is approximately:

\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)

\Delta V \approx 1005.310\,ft^{3}

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

7 0
3 years ago
Please help!!! I've been trying to figure this out all day!
MrRa [10]

Answer:

.55(220-a) + .90(220-a)

Step-by-step explanation:

3 0
3 years ago
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