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3 years ago

.A retailer sold two articles at same price. He sold the first one at 10% profit and the

Mathematics

1 answer:

IRISSAK [1]3 years ago

3 0

Step-by-step explanation:

so let's assume they both sold for 100 each the first article sold for 10% laws there by cost of article is 111.11 rounded the second article made 25% profit so cost must be 80 total cost is 8.89 or 5% profit I'm so sorry if im wrong

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What’s the pattern of -1, -8, -27, and -64

AURORKA [14]

Hello from MrBillDoesMath!

Answer:

a(n) = (-n)^3 where n = 1,2,3,...

Discussion:

The pattern 1,8,27, 64... is immediately recognizable as the the cube of the positive integers. But this question has a minus sign appearing before each entry, suggesting we try this:

- 1 = (-1)^3

-8 = (-2)^3

-27 = (-3)^3

-64 = (-4)^3

That's what the problem statement asked for . The answer is equivalently

-1 * (n^3)

Thank you,

MrB

6 0

3 years ago

Si f(x)=2x^{2}−3x+1 y g(x)=x−3. Entonces (f∘g)(−1) es:

WARRIOR [948]

Huh? this is very confusing 12

7 0

3 years ago

For the system of equations that follows, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in

Rainbow [258]

Answer:

$\begin{Vmatrix}1 & 0 & 0& 46/41 &180/41\\0 & 1 & 0&1/41 &-5/41\\0&0 & 1 & -8/41& -42/41\end{Vmatrix}$

Step-by-step explanation:

From the question we are told that

System of equations given as

x₁ + 3x₂ + x₃ + x₄ = 3;

2x₁ - 2x₂ + x₃ + 2x₄ = 8;

x₁ - 5x₂ + x₄ = 5

Matrix form

$\begin{Vmatrix}1 & 3 & 1&1&3\\2 & -2 & 1&2&8\\0&1 & -5 & 1& 5\end{Vmatrix}$ $\begin{Vmatrix}x_1\\x_2\\x_3\\x_4\end{Vmatrix}$

Generally the echelon reduction is mathematically applied as

$\begin{Vmatrix}1 & 3 & 1&1&3\\2 & -2 & 1&2&8\\0&1 & -5 & 1& 5\end{Vmatrix}$

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

$\begin{Vmatrix}1 & 3 & 1&1&3\\0 & -8 & -1&0&2\\0&1 & -5 & 1& 5\end{Vmatrix}$

Multiply the 2nd row by -1/8

$\begin{Vmatrix}1 & 3 & 1&1&3\\0 & 1 & 1/8&0&-1/4\\0&1 & -5 & 1& 5\end{Vmatrix}$

Add -1 times the 2nd row to the 3rd row

$\begin{Vmatrix}1 & 3 & 1&1&3\\0 & 1 & 1/8&0&-1/4\\0&0 & -41/8 & 1& 21/4\end{Vmatrix}$

Multiply the 3rd row by -8/41

$\begin{Vmatrix}1 & 3 & 1&1&3\\0 & 1 & 1/8&0&-1/4\\0&0 & 1 & -8/41& -42/41\end{Vmatrix}$

Add -1/8 times the 3rd row to the 2nd row

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

$\begin{Vmatrix}1 & 3 & 0&49/41&165/41\\0 & 1 & 0&1/41 &-5/41\\0&0 & 1 & -8/41& -42/41\end{Vmatrix}$

Add -3 times the 2nd row to the 1st row

$\begin{Vmatrix}1 & 0 & 0& 46/41 &180/41\\0 & 1 & 0&1/41 &-5/41\\0&0 & 1 & -8/41& -42/41\end{Vmatrix}$

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3 years ago

A pants factory pays it’s seamstresses by their production output if the company paid a seamstresses five cents for every four p

Oduvanchick [21]

P MO LP LP LP o look pop or pop pop o m op ok MN l opl

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3 years ago

Read 2 more answers

What number must you add to complete the square?<br> x^2 + 2x = -1

Andru [333]

$\large\begin{array}{l} \textsf{You must add 1 to complete the square.}\\\\ \textsf{There is a very common special product: (square of a sum)}\\\\ \mathsf{(a+b)^2=a^2+2ab+b^2}\\\\\\ \textsf{Take the given equation:}\\\\ \mathsf{x^2+2x=-1\qquad(i)}\\\\\\ \textsf{The first term is a square (x squared)}\\\\ \textsf{The second term is a product of 2, x and 1.} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{We need to find a proper value for b, so that if we add it to}\\\textsf{the left side of the equation, we get a perfect square.}\\\\ \textsf{So,}\\\\ \begin{array}{cccccccccc} \mathsf{a^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{a}&\!\!\!\cdot\!\!\!&\mathsf{b}&\!\!\!+\!\!\!&\mathsf{b^2}&=\mathsf{(a+b)^2}\\\\ \mathsf{\downarrow}&&&\!\!\!\!\!&\downarrow&\!\!\!\!\!&\downarrow&&\downarrow\\\\ \mathsf{x^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{x}&\!\!\!\cdot\!\!\!&\mathsf{1}&\!\!\!+\!\!\!&\mathsf{1^2}&=\mathsf{(x+1)^2} \end{array}\\\\\\ \textsf{Just by comparing the expressions above, we can conclude that}\\\textsf{b must be equal to 1, so we get a perfect square:}\\\\ \mathsf{(x+1)^2=x^2+2x+1} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{Back to (i), adding 1 to both sides:}\\\\ \mathsf{x^2+2x+1=-1+1}\\\\ \boxed{\begin{array}{c} \mathsf{(x+1)^2=0} \end{array}} \end{array}$

$\large\begin{array}{l}\\\\\\ \textsf{If you want to solve this equation for x, you will find}\\\\ \mathsf{x+1=0}\\\\ \mathsf{x=-1}\quad\longleftarrow\quad\textsf{(solution)} \end{array}$

If you're having problems understanding the answer, try seeing it through your browser: brainly.com/question/2112248

$\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best regards! :-)} \end{array}$

6 0

3 years ago

Read 2 more answers