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2 years ago

9

a and b together can do a work in 10 days B and C together can do the same work in 15 days a b and c together can do the same wo

rk in 8 days then how many days can B alone do this work

1 answer:

trasher [3.6K]2 years ago

4 0

Answer:

hauqvuauq shuavauvqsjahw w

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Can you help me solve systems of equations by elimination step by step<br> 6x+y=15<br> -7x-2y=-10

Anna71 [15]

Answer: (4, -9)

<u>Step-by-step explanation:</u>

Use elimination method. Manipulate one (or both) equations to eliminate one of the variables and solve for the remaining variable. <em>I will be eliminating y</em>

6x + y = 15 → 2(6x + y = 15) → 12x + 2y = 30

-7x - 2y = -10 → 1(-7x + 2y = -10) → <u> -7x - 2y = -10</u>

5x = 20

x = 4

Next, replace "x" with "4" into either equation and solve for y.

6(4) + y = 15

24 + y = 15

y = -9

<u>Check:</u>

Plug in x = 4 and y = -9 into the other equation to verify it makes a true statement.

-7x - 2y = -10

-7(4) - 2(-9) = -10

-28 - -18 = -10

-28 + 18 = -10

-10 = -10 $\checkmark$

6 0

2 years ago

GIVING BRAINLIEST!!!!!!!!<br><br> 1/2 x(6X4)+3+2<br><br> SHOW UR WORK PLS

mezya [45]

Answer: 1/2 times 6x equals 3x and then you do 1/2 times 4 and that equals 2 then multiply the 3 and 4 in the parenthesis and that is 12 then add 12 to the 3 and 2 and your answer is 17.

Step-by-step explanation:

7 0

2 years ago

A hoop, a uniform solid cylinder, a spherical shell, and a uniform solid sphere are released from rest at the top of an incline.

Serjik [45]

Answer:

Step-by-step explanation:

Given

Hoop, Uniform Solid Cylinder, Spherical shell and a uniform Solid sphere released from Rest from same height

Suppose they have same mass and radius

time Period is given by

$t=\sqrt{\frac{2h}{a}}$ ,where h=height of release

a=acceleration

$a=\frac{g\sin \theta }{1+\frac{I}{mr^2}}$

Where I=moment of inertia

a for hoop

$a=\frac{g\sin \theta }{1+\frac{mr^2}{mr^2}}$

$a=\frac{g\sin \theta }{2}$

a for Uniform solid cylinder

$a=\frac{g\sin \theta }{1+\frac{mr^2}{2mr^2}}$

$a=\frac{2g\sin \theta }{3}$

a for spherical shell

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{3mr^2}}$

$a=\frac{3g\sin \theta }{5}$

a for Uniform Solid

$a=\frac{g\sin \theta }{1+\frac{2mr^2}{5mr^2}}$

$a=\frac{5g\sin \theta }{7}$

time taken will be inversely proportional to the square root of acceleration

$t_1=k\sqrt{2}=1.414k$

$t_2=k\sqrt{\frac{3}{2}}=1.224k$

$t_3=k\sqrt{\frac{5}{3}}=1.2909k$

$t_4=k\sqrt{\frac{7}{5}}=1.183k$

thus first one to reach is Solid Sphere

second is Uniform solid cylinder

third is Spherical Shell

Fourth is hoop

3 0

2 years ago

Each square below represents one whole.

geniusboy [140]

Answer:

???69

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Help me please!!!! Please

Monica [59]

Answer:

5

Step-by-step explanation:

it's not necessary to round to the nearest tenth just simply half 10

3 0

2 years ago