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3 years ago

Evaluate each expression if x=6,y=8,z=3 2x+3y-z=

Mathematics

2 answers:

alexandr402 [8]3 years ago

7 0

12+24-3=33 (that's what you get when you plug each number into the correct value) so the answer is 33

Savatey [412]3 years ago

3 0

$x=6,\ \ y=8,\ \ z=3\\\\2x+3y-z=2\cdot 6+3\cdot 8-3=12+24-3=33$

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A store has clearance items that have been marked down by 60%. They are having a sale, advertising an additional 15% off clearan

astra-53 [7]

Answer: 34%

Step-by-step explanation:

Given

Price is marked down by 60% with an additional discount of 15% on the marked down price

Suppose x is the original price of the item

After marking down, it is

$\Rightarrow x-x\times 0.6=0.4x$

after additional discount it is

$\Rightarrow 0.4x(1-0.15)=0.4x\times 0.85=0.34x$

Effective discount on the item is

$\Rightarrow \dfrac{x-0.34x}{x}\times 100=66\%$

Thus, a person is paying $100-66=34\%$ of the original price

6 0

2 years ago

If a=b, then a-6=b-6

Dima020 [189]

Answer:

Step-by-step explanation:

it subtracts equally on each side

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:

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

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

Question 1!!! PLEASE HELP

andreev551 [17]

Quickly using a calculator (or long division) reveals that 191/238=<span>0.8025=80.25%, which is close to 80%=4/5
Similarly, 106/160=0.6625=66.25%, which is close to 66.67%=2/3</span>

7 0

3 years ago

the room to be covered is rectangular in shape with a width of 20 ft and a length of 28 ft .what is the area of room to be cover

kiruha [24]

A=L×W
A=20×28
A=560ft.

5 0

3 years ago

Read 2 more answers