Which expression is equivalent to -9x-1y-9/-15x5y-3?

On the 1st of January 2014, Carol invested some money in a bank account. The account pays 2.5% compound interest per year. On th

cluponka [151]

She started with x.
After 1 year, she had 1.025x.
She withdrew £1000, so now she has 1.025x - 1000.
Then it earned interest for 1 year and ended up as 1.025(1.025x - 1000).
The actual amount of money was £23 517.60.
Therefore,

1.025(1.025x - 1000) = 23517.60

1.025x - 1000 = 22 944

1.025x = 23 944

x = 23 360

Her original deposit was £23 360

5 0

3 years ago

Which polygon is always regular?

aleksley [76]

Square polygon is always regular, other have some scenarios which makes them regular but they are not always regular.

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 $ax_1+bx_2$ 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

Please help me with the problem

mixas84 [53]

Let's see what we're working on here

$\frac{1}{2} - 2(m) =?$
? = 0
$Simplify this →$ $\frac{1}{2}$

$\frac{1}2y} - 2(m) = ?$

$\frac{2(m)(2)}{2}$

$\frac{1 - 4(m)}{2} = ?$

$\frac{1 - 4(m)}{2} = 2$

$-4(m) + 1 = 0$
Subtract ( - )the number 1 from each of your sides on this part

4(m) = 1
Multiply( ×) the number -1 to your sides for this part of the equation

Therefore, the value of m is $\frac{1}{4}$
$m = \frac{1}{4}$

3 0

4 years ago

Read 2 more answers

Which of the following is an example of the identity property of 1?

Valentin [98]

Answer:

2 or 1

i mostly think its 2 but im not sure

Step-by-step explanation:

8 0

3 years ago