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Serggg [28]
3 years ago
6

How to get smarter in math?

Mathematics
2 answers:
anastassius [24]3 years ago
8 0
Practice, practice & practice.
frutty [35]3 years ago
3 0
You could arrange for a tutor.

You could ask your guardian to help explain.

You could ask your teacher.


Practice and study what you are having difficulty with.

Remember: PRACTICE MAKES PERFECT!!!

(not always but you will get better)
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
I need help with number 2 please.
Vedmedyk [2.9K]
The answer is 73.33333333 (repeated). You can also write it as 73 or 73.3 with a line over the 3 (after the decimal point) to indicate it being repeated.
8 0
3 years ago
What’s the password. Pleas help
Alecsey [184]

Password to what exactly?

4 0
3 years ago
The pyramid below has a square base.
zavuch27 [327]

The volume of the pyramid would be 2406.16 cubic cm.

<h3>How to find the volume of a square-based right pyramid?</h3>

Supposing that:

The length of the sides of the square base of the pyramid has = b units

The height of the considered square-based pyramid = h units,

The pyramid below has a square base.

h = 24.4 cm

b = 17.2 cm

Then, its volume is given by:

V = \dfrac{1}{3} \times b^2 \times h \: \rm unit^3

V = \dfrac{1}{3} \times 17.2^2 \times 24.4 \: \rm unit^3\\\\V = 2406.16

Therefore, the volume of the pyramid would be 2406.16 cubic cm.

Learn more about pyramid;

brainly.com/question/20324485

#SPJ1

3 0
2 years ago
Someone help me pleaseee!!
iren [92.7K]

Answer:

4 miles.

Step-by-step explanation:

3.50 ×4=11.50.

4 0
3 years ago
Read 2 more answers
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