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Fiesta28 [93]
3 years ago
11

-7y = -77 solve for y

Mathematics
2 answers:
babunello [35]3 years ago
5 0
Y=11
is the answer to your question! <3
Artist 52 [7]3 years ago
5 0

Answer:

y is 11

Step-by-step explanation:

-7y = -77

Divide both sides by -7

-7y/-7 = -77/-7

Since it's -77/-7, a negative divided by a negative gives us a positive output.

Y=11 is our final answer.

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What is the solution to the inequality x2 &lt; 16 – 6x?
zaharov [31]

Answer:

D

Step-by-step explanation:

2x < 16 - 6x

Bring the variables to one side

2x < 16 - 6x

+6x       +6x

8x < 16

Divide both sides by the coefficient

8x/8 < 16?8

x < 2

4 0
3 years ago
Mary earns $6 per hour as a waitress. Last week she took home her regular earnings of $6 per hour plus $185 in tips for a total
Daniel [21]
Mary worked 22 hours.


To solve first you would subtract the amount she made in tips from her total earnings (317-185)
= 132
Then you divide that answer by her hourly salary of $6 (132/6)
=22
7 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
15 a2 - 6ab- 8 a2+ 20 - 5ab - 31 + a2- ab
Ivanshal [37]

Step-by-step explanation:

Remove the parentheses: 15a²-6ab+8a²+20+5ab-31+a²-ab Combine like terms: 24a²-2ab-11

Answer: 24a²-2ab-11

4 0
3 years ago
Read 2 more answers
Water stations are placed every 600 meters of a fifteen kilometer race. How many water stations will be needed?
timurjin [86]

25 water stations, 15000/600 = 25

3 0
3 years ago
Read 2 more answers
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