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Leviafan [203]
3 years ago
6

Find the volume of the sphere ​

Mathematics
1 answer:
postnew [5]3 years ago
6 0
V = 4/3 r^3 hope this helps I’m not sure if it’s right
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What is the other point of the line is one is (-4,5) and the slope is -2/3
castortr0y [4]

let's notice, the "rise" is negative, meaning is going downwards.  Check the picture below.

6 0
3 years ago
A baseball team won 20 games and lost 10 games. What percent of the games did the team win?
omeli [17]
50% of the games were won
5 0
4 years ago
1 ) During his morning commute to work in rush hour traffic, Justin's average speed was 30 mi/h. During his afternoon commute ba
Hitman42 [59]

Answer:45 mi/h

Step-by-step explanation:

Recall: Speed = distance / time

Let the time taken for the morning journey be t_{1} and the time taken for the afternoon journey be t_{1} ,and the distance covered be d

that means for the first journey,

30 = \frac{d}{t_{1}}

d = 30 t_{1}

Also for the afternoon journey,

d = 60 t_{2}

Equating the two , since the same distance is being covered , we have

30 t_{1} = 60 t_{2}

that is

t_{1} = \frac{60t_{2}}{30}

t_{1} = 2 t_{2}

Also ,

total distance covered = 30 t_{1} + 60 t_{2}

Average speed = total distance / total time

 = 30 t_{1} + 60 t_{2} / t_{1} +t_{2}

  Recall that   t_{1} = 2 t_{2} , substitute this  into the formula for average speed , then we have

Average speed = 30(2t_{2}) +60 t_{2} / 3t_{2}

Average speed = 120t_{2} / 3t_{2}

Therefore :

Average speed = 40 mi/hr

                           

4 0
3 years ago
What is the value of b in the equation brainly (y^-9)^b=y^45?
riadik2000 [5.3K]

Answer:

its -5 just took the test

Step-by-step explanation:


4 0
4 years ago
Read 2 more answers
LINEAR ALGEBRA
kenny6666 [7]

Answer:

The value of the constant k so that \vec u_{3} is a linear combination of \vec u_{1} and \vec u_{2} is \frac{7}{10}.

Step-by-step explanation:

Let be \vec u_{1} = [2,3,1], \vec u_{2} = [4,1,0] and \vec u_{3} = [1, 2,k], \vec u_{3} is a linear combination of \vec u_{1} and \vec u_{3} if and only if:

\alpha_{1} \cdot \vec u_{1} + \alpha_{2} \cdot \vec u_{2} +\alpha_{3}\cdot \vec u_{3} = \vec O (Eq. 1)

Where:

\alpha_{1}, \alpha_{2}, \alpha_{3} - Scalar coefficients of linear combination, dimensionless.

By dividing each term by \alpha_{3}:

\lambda_{1}\cdot \vec u_{1} + \lambda_{2}\cdot \vec u_{3} = -\vec u_{3}

\vec u_{3}=-\lambda_{1}\cdot \vec u_{1}-\lambda_{2}\cdot \vec u_{2} (Eq. 2)

\vec O - Zero vector, dimensionless.

And all vectors are linearly independent, meaning that at least one coefficient must be different from zero. Now we expand (Eq. 2) by direct substitution and simplify the resulting expression:

[1,2,k] = -\lambda_{1}\cdot [2,3,1]-\lambda_{2}\cdot [4,1,0]

[1,2,k] = [-2\cdot\lambda_{1},-3\cdot \lambda_{1},-\lambda_{1}]+[-4\cdot \lambda_{2},-\lambda_{2},0]

[0,0,0] = [-2\cdot \lambda_{1},-3\cdot \lambda_{1},-\lambda_{1}]+[-4\cdot \lambda_{2},-\lambda_{2},0]+[-1,-2,-k]

[-2\cdot \lambda_{1}-4\cdot \lambda_{2}-1,-3\cdot \lambda_{1}-\lambda_{2}-2,-\lambda_{1}-k] =[0,0,0]

The following system of linear equations is obtained:

-2\cdot \lambda_{1}-4\cdot \lambda_{2}= 1 (Eq. 3)

-3\cdot \lambda_{1}-\lambda_{2}= 2 (Eq. 4)

-\lambda_{1}-k = 0 (Eq. 5)

The solution of this system is:

\lambda_{1} = -\frac{7}{10}, \lambda_{2} = \frac{1}{10}, k = \frac{7}{10}

The value of the constant k so that \vec u_{3} is a linear combination of \vec u_{1} and \vec u_{2} is \frac{7}{10}.

4 0
4 years ago
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