Explicit formula for<br><br> 1,5,3,6,12

4 and 5/6 - 1 and 5/6 ? help me pls ! i love you muah i mean

Otrada [13]

$4 \frac{5}{6} -1 \frac{5}{6}\\
\frac{29}{6}-\frac{11}{6}\\
\frac{18}{6}\\\\
3$

8 0

3 years ago

Read 2 more answers

Evaluate each logarithm. Do not use a calculator. ln ^3 square root e^4

Alona [7]

Answer:

$\large\boxed{\ln\sqrt[3]{e^4}=\dfrac{4}{3}}$

Step-by-step explanation:

$\text{Use}\\\\\sqrt[n]{a^m}=a^\frac{m}{n}\\\\\ln a^n=n\ln a\\\\\ln e=1\\-----------\\\\\ln\sqrt[3]{e^4}=\ln e^\frac{4}{3}=\dfrac{4}{3}\ln e=\dfrac{4}{3}\cdot1=\dfrac{4}{3}$

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 $ax_1+bx_2$ is 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

Which postulate (if any) should be used to show the triangles are congruent?

dezoksy [38]

Answer:

ASA

Step-by-step explanation:

SAS, it must go in order and SAS covers it up, Side, Angle, Side

7 0

2 years ago

What is the answer for 5-2+7x6-2?

sergiy2304 [10]

Answer: 43

Step-by-step explanation:

5-2+7x6-2

3+42-2

45-2

43

8 0

3 years ago

Explicit formula for 1,5,3,6,12