Answer:
Option C is correct.
Step-by-step explanation:
y = x^2-x-3 eq(1)
y = -3x + 5 eq(2)
We can solve by substituting the value of y in eq(2) in the eq(1)
-3x+5 = x^2-x-3
x^2-x+3x-3-5=0
x^2+2x-8=0
Now factorizing the above equation
x^2+4x-2x-8=0
x(x+4)-2(x+4)=0
(x-2)(x+4)=0
(x-2)=0 and (x+4)=0
x=2 and x=-4
Now finding the value of y by placing value of x in the above eq(2)
put x =2
y = -3x + 5
y = -3(2) + 5
y = -6+5
y = -1
Now, put x = -4
y = -3x + 5
y = -3(-4) + 5
y = 12+5
y =17
so, when x=2, y =-1 and x=-4 y=17
(2,-1) and (-4,17) is the solution.
So, Option C is correct.
Answer:
- True for Co-Prime Numbers
- False for Non Co-Prime Numbers
Step-by-step explanation:
<u>STATEMENT:</u> The LCM of two numbers is the product of the two numbers.
This statement is not true except if the two numbers are co-prime numbers.
Two integers a and b are said to be co-prime if the only positive integer that divides both of them is 1.
<u>Example: </u>
- Given the numbers 4 and 7, the only integer that divides them is 1, therefore they are co-prime numbers and their LCM is their product 28.
- However, consider the number 4 and 8. 1,2 and 4 divides both numbers, they are not co-prime, Their LCM is 8 which is not the product of the numbers.
Answer:

<u>Or in Decimal:</u>

Step-by-step explanation:
<u>Given expression :-</u>

<u>Solution :-</u>

This arithmetic expression may be rewritten as ;

Step 1 : <u>Cancel the zero of 10 and one zero of 100</u> :-

<em>Results to;</em>


Step 2: <u>Cancel 26 and 10</u><u> </u><u>by 2</u> :-

<em>Results to;</em>


<em>It can also be in Decimal.</em>
That is;

Hence, the answer of the expression would be 13/5 or 2.6 .

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