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tiny-mole [99]
3 years ago
5

How much acetic acid is in a 5-liter container that is marked 80% acetic acid? How

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
8 0

Answer:

1 liter I think?

Step-by-step explanation:

I don't know for sure but I am thinking it is 1 liter

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What is -1/2 + n = -5/8 what does n equal
Stels [109]

Answer: n=0.4 or 2/5

Step-by-step explanation:

4 0
3 years ago
An open rectangular box with square base has a volume of 256 cubic inches. determine the equation for the volume and surface are
Serjik [45]
Let
b-----------> the length side of the square box
h------------> the height of the box
SA---------> surface area of the box

we know that
[volume of the box]=b²*h
volume=256 in³
b²*h=256-------> h=256/b²-----> equation 1

surface area of the box=area of the base+perimeter of base*height

area of the base=b²
perimeter of the base=4*b

surface area=b²+(4*b)*h------> SA=b²+4*b*h-----> equation 2

substitute equation 1 in equation 2
SA=b²+4*b*[256/b²]-----> SA=b²+1024/b-----> SA=(b³+1024)/b

the answer is

the formula of the volume of the box is  V=b²*h-----> 256=b²*h

the formula of the surface area of the box are
SA=b²+4*b*h
SA=(b³+1024)/b
4 0
3 years ago
Show your work and solve please!
bixtya [17]

Answer:

y = 3

Step-by-step explanation:

-5y + 8 = -7

subtract 8 from both sides

-5y = -15

now, divide -5 from both sides

y = 3

4 0
2 years ago
Read 2 more answers
Line that passes through (-2,1) and (4,13)
Delvig [45]
Whats the question??

6 0
3 years ago
What degree of rotation is represented on this matrix
Korvikt [17]

Answer:

Option B is correct

the degree of rotation is, -90^{\circ}

Step-by-step explanation:

A rotation matrix is a matrix that is used to perform a rotation in Euclidean space.

To find the degree of rotation using a standard rotation matrix i.e,

R = \begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}

Given the matrix: \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

Now, equate the given matrix with standard matrix we have;

\begin{bmatrix}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} =  \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix}

On comparing we get;

\cos \theta = 0       and -\sin \theta =1  

As,we know:

  • \cos \theta = \cos(-\theta)
  • \sin(-\theta) = -\sin \theta

\cos \theta = \cos(90^{\circ}) = \cos( -90^{\circ})

we get;

\theta = -90^{\circ}

and

\sin \theta =- \sin (90^{\circ}) = \sin ( -90^{\circ})

we get;

\theta = -90^{\circ}

Therefore, the degree of rotation is, -90^{\circ}

7 0
3 years ago
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