If you mean |2-2| then the absolute value is 0
If you are talking about |2| |-2| then the absolute values are 2 and 2
Q11
Suppose the two numbers are x and y so xy = 3000
the HCF = 10
we know that product of two numbers = product of their HCF and LCM
XY = HCF * LCM
3000 = 10 * LCM
3000/10 = LCM
300 = LCM
Answer 300
Q12 : Let us say second number is x so
product of x and 160 = product of HCF and LCM of ( x and 160 )
160 x = 32 * 1760
x= 32 * 1760 / 160
x= 352
Answer : 352
Answer:
The expression 6x should be 6x^2.
Step-by-step explanation:
Given is the chart of multiplication of the binomial by the trinomial.
Let's check each element in the chart:-
3x times x^2 = 3x^3
3x times 2x = 6x^2
3x times 4 = 12x
2 times x^2 = 2x^2
2 times 2x = 4x
2 times 4 = 8
From the given chart, we can identify that <u>6x should be 6x^2.</u>
Hence. option B is correct, i.e. The expression 6x should be 6x^2.
Explicit Functiony = f(x) is said to define y explicitly as a function of x because the variable y appears alone on one side of the equation and does not appear at all on the other side. (ex. y = -3x + 5)Implicit FunctionAn equation in which y is not alone on one side. (ex. 3x + y = 5)Implicit DifferentiationGiven a relation of x and y, find dy/dx algebraically.d/dx ln(x)1/xd/dx logb(x) (base b)1/xln(b)d/dx ln(u)1/u × du/dxd/dx logb(u) (base b)1/uln(b) × du/dx(f⁻¹)'(x) = 1/(f'(f⁻¹(x))) iff is a differentiable and one-to-one functiondy/dx = 1/(dx/dy) ify = is a differentiable and one-to-one functiond/dx (b∧x)b∧x × ln(b)d/dx e∧xe∧xd/dx (b∧u)b∧u × ln(b) du/dxd/dx (e∧u)e∧u du/dxDerivatives of inverse trig functionsStrategy for Solving Related Rates Problems<span>1. Assign letters to all quantities that vary with time and any others that seem relevant to the problem. Give a definition for each letter.
2. Identify the rates of change that are known and the rate of change that is to be found. Interpret each rate as a derivative.
3. Find an equation that relates the variables whose rates of change were identified in Step 2. To do this, it will often be helpful to draw an appropriately labeled figure that illustrates the relationship.
4. Differentiate both sides of the equation obtained in Step 3 with respect to time to produce a relationship between the known rates of change and the unknown rate of change.
5. After completing Step 4, substitute all known values for the rates of change and the variables, and then solve for the unknown rate of change.</span>Local Linear Approximation formula<span>f(x) ≈ f(x₀) + f'(x₀)(x - x₀)
f(x₀ + ∆x) ≈ f(x₀) + f'(x₀)∆x when ∆x = x - x₀</span>Local Linear Approximation from the Differential Point of View∆y ≈ f'(x)dx = dyError Propagation Variables<span>x₀ is the exact value of the quantity being measured
y₀ = f(x₀) is the exact value of the quantity being computed
x is the measured value of x₀
y = f(x) is the computed value of y</span>L'Hopital's RuleApplying L'Hopital's Rule<span>1. Check that the limit of f(x)/g(x) is an indeterminate form of type 0/0.
2. Differentiate f and g separately.
3. Find the limit of f'(x)/g'(x). If the limit is finite, +∞, or -∞, then it is equal to the limit of f(x)/g(x).</span>
