Question

HELP me ïñ this question please​

Answer 1

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Let the fixed charge and the charge for each extra day be ‘a’and ‘b’ respectively.

Let the fixed charge and the charge for each extra day be ‘a’and ‘b’ respectively. Given, a lending library has a fixed charge for the first three days and an additional charge for each day thereafter Saritha paid Rs 27 for a book kept for seven days

⇒ a + 4b = 27 -------- (1)

Susy paid Rs 21 for the book she kept for five days

⇒ a + 2b = 21 --------- (2)

Subtracting eq2 from eq1

⇒ 2b = 6

⇒ b = 3

Putting this value in eq(1), we get

⇒ a + 4(3) = 2

⇒ a = 27 - 12

⇒ a = 15

Therefore, fixed charge,

a = 15

Rupees and charge thereafter,

b = 3

Rupees per day

⇒ a = Rs. 15

Answer 2

Step-by-step explanation:

Given Equation :

\begin{gathered}\sf { 15 \Big [ 4 + \Big ( -2 \Big )\Big ] = \Big [ \Big ( 15 \times 4 \Big ) \Big] + \Big [15 \times \Big ( -2 \Big ) \Big ] }\\\end{gathered}

15[4+(−2)]=[(15×4)]+[15×(−2)]

To find :

Verify the equation i.e, either it satisfies the distributive property or not.

Clarification :

Here, we are given a equation. We are asked to verify the equation i.e, either it satisfies the distributive property or not. Firstly let's understand what exactly the distributive property is.

Multiplication is distributive over addition and subtraction. That means, if a, b and c are rational numbers then :

a( b + c ) = ab + bc

a( b - c ) = ab - bc

Explication of steps :

LHS :

\begin{gathered}\longrightarrow \sf { 15 \Big [ 4 + \Big ( -2 \Big ) \Big ]}\\\end{gathered}

⟶15[4+(−2)]

\begin{gathered}\longrightarrow \sf { 15 \Big [ 4 - 2 \Big ]}\\\end{gathered}

⟶15[4−2]

\begin{gathered}\longrightarrow \sf { 15 \Big [ 2\Big ]}\\\end{gathered}

⟶15[2]

\begin{gathered}\longrightarrow \boxed{\sf { 30}}\\\end{gathered}

⟶

30

RHS :

\begin{gathered}\longrightarrow \sf { \Big [ \Big ( 15 \times 4 \Big ) \Big] + \Big [ 15 \times \Big ( -2 \Big ) \Big ] }\\\end{gathered}

⟶[(15×4)]+[15×(−2)]

\begin{gathered}\longrightarrow \sf { \Big [ 60 \Big] + \Big [ -30\Big ] }\\\end{gathered}

⟶[60]+[−30]

\begin{gathered}\longrightarrow \sf { 60 - 30 }\\\end{gathered}

⟶60−30

\begin{gathered}\longrightarrow \boxed{\sf { 30}}\\\end{gathered}

⟶

30

Comparing LHS and RHS, we get :

L.H.S = R.H.S

Henceforth, verified!