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agasfer [191]
3 years ago
7

Can anyone please help me with these questions please. thanks.

Mathematics
2 answers:
lisov135 [29]3 years ago
7 0
5 is 3\4 be you count one two three
Pani-rosa [81]3 years ago
3 0
1.) 3/4
2.) It snowed 1/4 more on Tuesday than it did on Monday.
You might be interested in
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
In a standard normal distribution 95% of the data is within +/- _______standard deviations of the mean.
raketka [301]
Your answer is 2 okay
8 0
3 years ago
Read 2 more answers
Can anyone answer this question please<br> 6[y-2]/7-12=2[y-7]/3
Kaylis [27]

Answer:

y =47.5.

Step-by-step explanation:

First eliminate the fractions by multiplying through by the LCM of 7 and 3 which is 21:

21* 6[y-2]/7-21*12 = 21*2[y-7]/3

18(y - 2) - 252 = 14(y - 7)

18y -36 - 252 = 14y - 98

18y - 14y = -98 + 36 + 252

4y = 190

y = 190/4

y = 47.5.

8 0
2 years ago
−2(x+1/4)+1=5 PLEASE HELP
PilotLPTM [1.2K]

Answer: x=−9/4

Step-by-step explanation:

Let's solve your equation step-by-step.

−2(x+14)+1=5

Step 1: Simplify both sides of the equation.

−2(x+14)+1=5(−2)(x)+(−2)(14)+1=5(Distribute)−2x+

−1

2

+1=5

(−2x)+(

−1

2

+1)=5(Combine Like Terms)

−2x+

1

2

=5

−2x+

1

2

=5

Step 2: Subtract 1/2 from both sides.

−2x+

1

2

−

1

2

=5−

1

2

−2x=

9

2

Step 3: Divide both sides by -2.

−2x

−2

=

9

2

−2

x=

−9

4

7 0
2 years ago
Read 2 more answers
Disssss tooo helllllllpppp shoals shzn
iren2701 [21]

Answer:

3/2

Step-by-step explanation:

Use this formula y2-y1/x2-x1

PLUG IN YOUR POINTS

-3 - -9/ 8- 4 =

6/4 (simplify)

1.5 or 3/2

Hope this helps ya!!

7 0
3 years ago
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