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katrin2010 [14]
3 years ago
11

the total cost of the trio is $600 what is the compatable number to estimate the total cost of the field trip

Mathematics
1 answer:
saveliy_v [14]3 years ago
3 0
Are there any other things that cost money that are in the equation?
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Find f'(x) <br>f(x)=(1-2x^2)^3
Hunter-Best [27]

Answer:  f'(x) = 12x  + 48x³ - 48x⁵

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

f(x) = (1 - 2x²)³

     = (1 - 2x²)(1 - 4x² + 4x⁴)

         1 - 4x² + 4x⁴

        <u>    -2x² + 8x⁴ - 8x⁶</u>

     =  1 - 6x² + 12x⁴ - 8x⁶

f'(x) = 0 -2(6)x²⁻¹ + 4(12)x⁴⁻¹ - 6(8)x⁶⁻¹

      =     12x  + 48x³ - 48x⁵

3 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
Last month you went on three hikes. one hike was2 3/4 miles long one was 1 1/2 miles long and was 3 3/4 miles long. in decimal h
creativ13 [48]
2 3/4 = 2.75

1 1/2 = 1.5 

3 3.4 = 3.75

2.75 + 1.5 + 3.75 = 8

You hiked 8 miles last month in total.
8 0
3 years ago
Read 2 more answers
PLEASE HELP ME
myrzilka [38]
Mike's withdrawals will be treated as negative numbers.

Since Mike withdrew $60 three times, use the following equation to find the change in his balance:

-60 * 3 = -180

The change in Mike's account balance will be -$180.
6 0
3 years ago
Read 2 more answers
What place value is 7 in 7,000.2
Dennis_Churaev [7]

Answer:

thousands place

Step-by-step explanation:

Have a nice day

8 0
3 years ago
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