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2 years ago

What is 28 % as a fraction?

Mathematics

2 answers:

Aleksandr [31]2 years ago

8 0

28/100 And that can be simplified 14/50 to 7/25 So 7/25 is the most simplified version

IgorLugansk [536]2 years ago

5 0

The answer is 7/25 simplified and if you dont want it to be simplified, it is 28/100

You might be interested in

BRAINLIEST PLUS 30 POINTS

aleksklad [387]

umm I think it =(m)(x+−x)

=(m)(x)+(m)(−x)

=mx−mx

hope it's right and you can download mathpapa

7 0

2 years ago

A shop sells fence panels with a wooden frame going all the way round each panel. The price of each panel is based on the area o

vagabundo [1.1K]

Answer:

a) Each 1m² of fence panel costs: £2

b) I metre of wooden frame costs: £1

Step-by-step explanation:

Let the price of 1 sq.m. of fence panel be x

Let the price of 1 m of wooden frame be y

In Fig 1

Length = 4 m

Breadth = 3 m

Area = $Length \times breadth = 4 \times 3 = 12 sq.m.$

Cost of 1 sq.m. = x

Cost of 12 sq.m. = 12x

Perimeter =2(l+b)=2(4+3)=14 m

Cost of 1 m of wooden frame = y

Cost of 14 m of wooden frame = 14y

We are given that total cost of framing and fencing is 38

So. 12x+14y=38 ----1

In Fig 2

Length = 6 m

Breadth = 2 m

Area = $Length \times breadth = 6 \times 2 = 12 sq.m$ .

Cost of 1 sq.m. = x

Cost of 12 sq.m. = 12x

Perimeter =2(l+b)=2(6+2)=16 m

Cost of 1 m of wooden frame = y

Cost of 16 m of wooden frame = 16y

We are given that total cost of framing and fencing is 40

So. 12x+16y=40---2

Substitute the value of 12x form 1 in 2

38-14y+16y=40

2y=2

y=1

Substitute the value of y in 2

12x+16=40

12x=24

x=2

So, cost of 1 sq.m. of fence panel is 2 and cost of 1 m of wooden frame is 1

In fig 3

To verify :

Length = 3 m

Breadth = 5 m

Area = $Length \times breadth = 3 \times 5 = 15 sq.m.$

Cost of 1 sq.m. = 2

Cost of 12 sq.m. = $15 \times 2 = 30$

Perimeter =2(l+b)=2(3+5)=16 m

Cost of 1 m of wooden frame = 1

Cost of 16 m of wooden frame = 16

So, Total cost = 30+16=46

We are given that cost of fencing and framing is 46

Thus verified

Hence

a) Each 1m² of fence panel costs: £2

b) I metre of wooden frame costs: £1

5 0

3 years ago

Convert the given system of equations to matrix form

yuradex [85]

Answer:

The matrix form of the system of equations is $\left[\begin{array}{ccccc}1&1&1&1&-3\\1&-1&-2&1&2\\2&0&1&-1&1\end{array}\right] \left[\begin{array}{c}x&y&w&z&u\end{array}\right] =\left[\begin{array}{c}5&4&3\end{array}\right]$

The reduced row echelon form is $\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

The vector form of the general solution for this system is $\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

Step-by-step explanation:

Convert the given system of equations to matrix form

We have the following system of linear equations:

$x+y+w+z-3u=5\\x-y-2w+z+2u=4\\2x+w-z+u=3$

To arrange this system in matrix form (Ax = b), we need the coefficient matrix (A), the variable matrix (x), and the constant matrix (b).

$A= \left[\begin{array}{ccccc}1&1&1&1&-3\\1&-1&-2&1&2\\2&0&1&-1&1\end{array}\right]$

$x=\left[\begin{array}{c}x&y&w&z&u\end{array}\right]$

$b=\left[\begin{array}{c}5&4&3\end{array}\right]$

Use row operations to put the augmented matrix in echelon form.

An augmented matrix for a system of equations is the matrix obtained by appending the columns of b to the right of those of A.

So for our system the augmented matrix is:

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\1&-1&-2&1&2&4\\2&0&1&-1&1&3\end{array}\right]$

To transform the augmented matrix to reduced row echelon form we need to follow this row operations:

add -1 times the 1st row to the 2nd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&-2&-3&0&5&-1\\2&0&1&-1&1&3\end{array}\right]$

add -2 times the 1st row to the 3rd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&-2&-3&0&5&-1\\0&-2&-1&-3&7&-7\end{array}\right]$

multiply the 2nd row by -1/2

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&-2&-1&-3&7&-7\end{array}\right]$

add 2 times the 2nd row to the 3rd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&0&2&-3&2&-6\end{array}\right]$

multiply the 3rd row by 1/2

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&3/2&0&-5/2&1/2\\0&0&1&-3/2&1&-3\end{array}\right]$

add -3/2 times the 3rd row to the 2nd row

$\left[\begin{array}{ccccc|c}1&1&1&1&-3&5\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

add -1 times the 3rd row to the 1st row

$\left[\begin{array}{ccccc|c}1&1&0&5/2&-4&8\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

add -1 times the 2nd row to the 1st row

$\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

Find the solutions set and put in vector form.

Interpret the reduced row echelon form:

The reduced row echelon form of the augmented matrix is

$\left[\begin{array}{ccccc|c}1&0&0&1/4&0&3\\0&1&0&9/4&-4&5\\0&0&1&-3/2&1&-3\end{array}\right]$

which corresponds to the system:

$x+1/4\cdot z=3\\y+9/4\cdot z-4u=5\\w-3/2\cdot z+u=-3$

We can solve for z:

$z=\frac{2}{3}(u+w+3)$

and replace this value into the other two equations

$x+1/4 \cdot (\frac{2}{3}(u+w+3))=3\\x=-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}$

$y+9/4 \cdot (\frac{2}{3}(u+w+3))-4u=5\\y=\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}$

No equation of this system has a form zero = nonzero; Therefore, the system is consistent. The system has infinitely many solutions:

$x=-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}\\y=\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}\\z=\frac{2u}{3}+\frac{2w}{3}+2$

where u and w are free variables.

We put all 5 variables into a column vector, in order, x,y,w,z,u

$x=\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=\left[\begin{array}{c}-\frac{u}{6} -\frac{w}{6}+\frac{5}{2}&\frac{5u}{2}-\frac{3w}{2}+\frac{1}{2}&w&\frac{2u}{3}+\frac{2w}{3}+2&u\end{array}\right]$

Next we break it up into 3 vectors, the one with all u's, the one with all w's and the one with all constants:

$\left[\begin{array}{c}-\frac{u}{6}&\frac{5u}{2}&0&\frac{2u}{3}&u\end{array}\right]+\left[\begin{array}{c}-\frac{w}{6}&-\frac{3w}{2}&w&\frac{2w}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

Next we factor u out of the first vector and w out of the second:

$u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

The vector form of the general solution is

$\left[\begin{array}{c}x&y&w&z&u\end{array}\right]=u\left[\begin{array}{c}-\frac{1}{6}&\frac{5}{2}&0&\frac{2}{3}&1\end{array}\right]+w\left[\begin{array}{c}-\frac{1}{6}&-\frac{3}{2}&1&\frac{2}{3}&0\end{array}\right]+\left[\begin{array}{c}\frac{5}{2}&\frac{1}{2}&0&2&0\end{array}\right]$

7 0

3 years ago

Which statements about the area of the faces of the rectangular prism are true? Select all that apply.

LenaWriter [7]

Answer: The front face has an area of 30 ft2. , The top face has an area of 40 ft2. ,

Step-by-step explanation: SOrry if im wrong

5 0

2 years ago

What is the range of the function graphed below?

Xelga [282]

<h2>Answer:</h2>

Option: B is the correct answer.

The range of the function is:

B. 5 < y < ∞

<h2>Step-by-step explanation:</h2>

Range of a function--

The range of a function is the set of all the values that is attained by the function.

By looking at the graph of the function we see that the function tends to 5 when x→ -∞ and the function tends to infinity when x →∞

Also, the function is a strictly increasing function.

This means that the function takes every real value between 5 and ∞ .

i.e. The range of the function is: (5,∞)

Hence, the answer is:

Option: B

8 0

3 years ago