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3 years ago

Solve the inequality for x. 17 < x - 10 Simplify your answer as much as possible.

Mathematics

2 answers:

Sonbull [250]3 years ago

8 0

Step-by-step explanation:

x = 27

Can i have brainly please

natali 33 [55]3 years ago

3 0

Answer:

27<x

Step-by-step explanation:

you would add 10 to both sides to isolate the x

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George earned a total of $342 in simple interest from two separate accounts. In an account earning 5% interest, George invested

Maslowich

Answer: he invested $6000 in the account earning 5% interest and $2100 in the other account earning 2% interest

Step-by-step explanation:

Let x represent the amount invested in the account earning 5% interest.

Let y represent the amount invested in the account earning 2% interest.

In an account earning 5% interest, George invested $1800 more than twice the amount he invested in an account earning 2%. It means that

x = 2y + 1800

The formula for simple interest is expressed as

I = PRT/100

Where

P represents the principal

R represents interest rate

T represents time in years

I = interest after t years

Assuming the duration for both investments is 1 year,

The interest on the first account would be

I = (x × 5 × 1)/100 = 0.05x

The interest on the second account would be

I = (y × 2 × 1)/100 = 0.02y

George earned a total of $342 in simple interest from two separate accounts. This means that

0.05x + 0.02y = 342 - - - - - - - - - - 1

Substituting x = 2y + 1800 into equation 1, it becomes

0.05(2y + 1800) + 0.02y = 342

0.1y + 90 + 0.02y = 342

0.1y + 0.02y = 342 - 90

0.12y = 252

y = 252/0.12 = 2100

x = 2y + 1800 = 2 × 2100 + 1800 = $6000

4 0

3 years ago

Can you please answer number 82

hichkok12 [17]

A.) the answer is 2x+y+7
b.)the answer is 2x+7y-9
c.)the answer is 24x+48
d.)the answer is 24x+6
e.)the answer is 5x+9

8 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:

$\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon F \\\\\to a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))$