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Fed [463]
3 years ago
5

Solve for x in both!!!!!

Mathematics
2 answers:
inna [77]3 years ago
8 0

Answer:

1. No solution

2. x = 0

Step-by-step explanation:

-2(x-2)-4x=3(x+1)-9x

-2x + 4 - 4x = 3x + 3 - 9x

-6x + 4 = 3 - 6x

0 = -1

—

5(x+2)-3=3x-8x+7

5x + 10 -3 = 3x - 8x +7

5x + 7 = 5x + 7

x = 0

zvonat [6]3 years ago
4 0

Answer:

- 2(x - 2) - 4x = 3(x + 1) - 9x \\  - 2x + 4 - 4x = 3x + 3 - 9x \\ 9x - 3x - 2x - 4x = 3 - 4 \\ 9x - 9x =  - 1 \\ 0x =  - 1 \\ 0 =  - 1

So ; No solution

___o___o____

5(x + 2) - 3 = 3x - 8x + 7 \\ 5x + 10 - 3 =  - 5x + 7 \\ 5x + 7 =  - 5x + 7 \\ 5x + 5x = 7 - 7 \\ 10x = 0 \\  \\ x =  \frac{0}{10}  \\  \\ x = 0

I hope I helped you^_^

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Evaluate A/B for a = 1/2 and b = -3/7
bonufazy [111]

Answer:

-7/6

Step-by-step explanation:

If a = 1/2 and b = -3/7, then your given:

1/2 divided by -3/7=

-7/2*3=

-7/6

Sorry if its a bit unclears

4 0
3 years ago
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Find area of this figure 3 ft 4 ft 5ft
Ahat [919]

Answer:

26.14

Step-by-step explanation:

For the triangles: 3*4=12

For the circle: 28.27/2=14.135

12+14.135=26.135

~26.14

6 0
3 years ago
In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank
fredd [130]

Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) Amount \ of \, water \ remaining \ in \, the \ tank \ is \  \frac{x^3(6-\pi) }{6}

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The  volume of the sphere = \frac{4}{3} \pi r^3

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = \frac{4}{3} \pi \frac{x^3}{8}= \frac{1}{6} \pi x^3

For tank B

Volume of tank = x³

The  volume of the spheres = 8 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = 8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}

For tank C

Volume of tank = x³

The  volume of the spheres = 64 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 4·D = x therefore;

r = x/8 and the volume of the sphere is thus;

volume of the spheres = 64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The  volume of the spheres = 512 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = 512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}

Amount \ of \ water \, remaining \, in \, the \ tank =  \frac{x^3(6-\pi) }{6}.

5 0
3 years ago
The boeing 747 airplane is 250 feet in length. If a model of this plane is 8 inches long , what scale was used to create this mo
cluponka [151]

Answer: The wingspan is 36 m.

Using a proportion to solve this, we write the scale factor as a ratio first:  19/38, since 19 is the size of the model's length and 38 is the real length.  For the second ratio, we have 18 for the size of the model wingspan and x for the real wingspan:

19/38 = 18/x

Cross multiply:

19*x = 38*18

19x = 684

8 0
3 years ago
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Dvinal [7]

Answer:

1st solution is -2x+12

2nd solution is 5x+31

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Step-by-step explanation:

4 0
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