Answer:
The half life of the car is 3.98 years.
Step-by-step explanation:
The value of the car after t years is given by the following equation:
In which V(0) is the initial value and r is the constant decay rate, as a decimal.
The value of a certain car decreases by 16% each year.
This means that 
So
What is the 1⁄2-life of the car?
This is t for which V(t) = 0.5V(0). So
The half life of the car is 3.98 years.
Hi,
To solve this problem, Let us take the LCM of 10 and 16 which will come 80.
Now suppose the cost price of 10 tables =₹n CP of 80 tables will be ₹ 8n
According to the question, CP of 10 tables is equal to the SP of 16 tables, then
the SP of 16 tables will also be ₹ n.
So, SP of 80 tables will be ₹ 5n
So, Loss = CP-SP
→ 8n - 5n = ₹ 3n
Loss%= (3n×100)/8n
Loss%= 37.5%.
Hence the correct answer will be a <u>loss of 37.5%.</u>
Answer:
6 i think
Step-by-step explanation:
A
=
h
b
b
2
=
4
·
3
2
=
6
The factorization of the expression 36x⁴y³ - 16x³y is 4x³y(9xy² - 4)
<h3>Factorization</h3>
Factorization or factoring is defined as the breaking or decomposition of an entity (for example a number, a matrix, or a polynomial) into a product of another entity, or factors, which when multiplied together give the original number. Factorization of an algebraic expression means writing the given expression as a product of its factors. These factors can be numbers, variables, or an algebraic expression.
To the factor, a number means to break it up into numbers that can be multiplied to get the original number.
In the given expression, 36x⁴y³ - 16x³y can be factorized as
36x⁴y³ - 16x³y = 4x³y(9xy² - 4)
The common factor in the expression is 4x³y
Learn more on factorization here;
brainly.com/question/8021175
#SPJ1
Answer:
-15
Step-by-step explanation:
We proceed as follows;
In this question, we want to fill in the blank so that we can have the resulting expression expressed as the product of two different linear expressions.
Now, what to do here is that, when we factor the first two expressions, we need the same kind of expression to be present in the second bracket.
Thus, we have;
2a(b-3) + 5b + _
Now, putting -15 will give us the same expression in the first bracket and this gives us the following;
2a(b-3) + 5b-15
2a(b-3) + 5(b-3)
So we can have ; (2a+5)(b-3)
Hence the constant used is -15
