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svet-max [94.6K]

2 years ago

11

At noon, a tank contained 5 cm of water. After several hours, it contained 4 cm of water. What is the percent decrease of water

in the tank?

1 answer:

Radda [10]2 years ago

6 0

Answer:

the decrease percentage is 25%

You might be interested in

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

2 years ago

Please help due in 10 min.

sleet_krkn [62]

Answer:

A: equiangular S: Right

Step-by-step explanation:

5 0

2 years ago

Can someone please help me <br>a<br>show your work please

sertanlavr [38]

Answer:

x= 3

Step-by-step explanation:

19x-5=7+15x

19x-15x=7+5

4x=12

x=12/4

x=3

8 0

2 years ago

Read 2 more answers

Write the expression 5x+2x^2-7+2x^3

avanturin [10]

2x^3+2x^2+5x-7, That would be the answer.

3 0

2 years ago

What’s three points that satisfy the system of inequality’s of 4x+6y<24

Vlada [557]

Answer:

(0,0), (1,1), (2,2)

Step-by-step explanation:

When testing to find possible points in situations like this, I always start by testing with the origin point (0,0).

In this case:

4x+6y<24 ==> 0 + 0 < 24 TRUE, it satisfies the inequality.

We then try with (1,1):

4x+6y<24 ==> 4 + 6 < 24 TRUE, it satisfies the inequality.

And with (2,2):

4x+6y<24 ==> 8 + 12 < 24 TRUE, it satisfies the inequality.

8 0

2 years ago