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Snowcat [4.5K]
2 years ago
15

A ball kicked a distance of 20 meters on Earth would travel 2 13/20 times farther on Mars. On Phobos, the ball would travel appr

oximately 716.98 times farther than on Mars. About how many kilometers would the ball travel on Phobos? Show your work
Mathematics
1 answer:
Mademuasel [1]2 years ago
4 0

Answer: ask the teacher

Step-by-step explanation:

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Please answer the question
shusha [124]
The answer should be C. 3
6 0
3 years ago
Which of the following sets are subspaces of R3 ?
Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

\to A={(x,y,z)|3x+8y-5z=2} \\\\\to  for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                        =3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)

The set A is not part of the subspace R^3

for point B:

\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon  B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                             =-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0

\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon  B

The set B is part of the subspace R^3

for point C: \to C={(x,y,z)|x

In this, the scalar multiplication can't behold

\to for (-2,-1,2) \varepsilon  C

\to -1(-2,-1,2)= (2,1,-1) ∉ C

this inequality is not hold

The set C is not a part of the subspace R^3

for point D:

\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)

The scalar multiplication s is not to hold

\to for (-4, 1,2)\varepsilon  D\\\\\to  -1(-4,1,2) = (4,-1,-2) ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace R^3

For point E:

\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to  a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\

The  x_1, x_2 is the arbitrary, in which ax_1+bx_2is arbitrary  

\to a(x_1,0,0)+b(x_2,0,0) \varepsilon  E

The set E is the part of the subspace R^3

For point F:

\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon  F \\\\\to  a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))

The x_1, x_2 arbitrary so, they have ax_1+bx_2 as the arbitrary \to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F

The set F is the subspace of R^3

5 0
3 years ago
What are the coordinates of the vertex of the parabola y=x^2+2x-1
kozerog [31]
Y = x^2 + 2x - 1 = x^2 + 2x + 1 - 1 - 1 = (x + 1)^2 - 2
The vertex of a parabola given by y = a(x - h)^2 + k is (h, k).

Therefore, the vertex of y =  (x + 1)^2 - 2 is (-1, -2)
3 0
2 years ago
Factor Completely: 3x2 + 9x - 30
enyata [817]

Answer:

3(x-2)(x+5)

Step-by-step explanation:

1. Factor out common term 3

3(x^2 +3x-10)

2. Factor (x^2 +3x-10)

3(x-2)(x+5)

8 0
3 years ago
Solve for a a^2+17=42?
grandymaker [24]

We are given the following equation:

a^2+17=42

Subtract both sides by 17

a^2=25

Take the positive/negative square root of both sides

a=\pm5

This should be your answer. Let me know if you have any questions, thanks!

6 0
3 years ago
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