Size 40 is the cheapest, because the formula to find lowest price of unit is dividing the price by the unit. so 5.49/24≈ 0.23, 8.00/40=0.2, 15.99/64≈ 0.25
Power and chain rule (where the power rule kicks in because
):

Simplify the leading term as

Quotient rule:

Chain rule:


Put everything together and simplify:







Answer:
<h3>
Acute Angles: ∠TLS, ∠SLT, ∠ULR</h3><h3>
Right Angles: ---------</h3><h3>
Obtuse Angles: ∠RLT, ∠SLU, ∠ULS,</h3><h3>
Straight Angles: ∠RLS, ∠TLU </h3><h3>
Not angles: ∠TRL </h3>
Step-by-step explanation:
The lines intersect at point L, so all angles have a vertex (middle letter) L so there is no angle TRL
Straight angle is a line with dot-vertex, so the straight angles are ∠RLS and ∠TLU.
∠TLS is less than 90° then it is acute angle (∠SLT is the same angle). ∠ULR is vertex angle to ∠TLS, so it's also acute angle.
Two angles adding to straight angle mean that they are both right angles or one is acute and the second is obtuse. ∠TLS is acute so ∠RLT is obtuse (they adding to ∠RLS) and ∠SLU is obtuse (they adding to ∠TLU). ∠ULS is the same angle as ∠SLU.
Chain rule
y=f(g(x))
y´=(d f(gx)/d g)(d g/d x)
or
y=y(v) and v=v(x), then dy /dx=(dy/dv)(dv/dx)
in our case:
y=sin (v)
v=arcsin(x)
dy/dv=d sin (v)/dv=cos (v)=cos(arcsin(x)
dv/dx=d arcsin(x)/dx=1/√(1-x²)
dy/dx=[cos (arcsin(x))]/√(1-x²)
Answer: d sin(arcsin(x))/dx=[cos (arcsin(x))]/√(1-x²)
Answer:
z = -10
General Formulas and Concepts:
<u>Pre-Algebra</u>
- Order of Operations: BPEMDAS
- Equality Properties
Step-by-step explanation:
<u>Step 1: Define equation</u>
-4 - (z - 6) = 12
<u>Step 2: Solve for </u><em><u>z</u></em>
- Distribute negative: -4 - z + 6 = 12
- Combine like terms: -z + 2 = 12
- Subtract 2 on both sides: -z = 10
- Divide -1 on both sides: z = -10
<u>Step 3: Check</u>
<em>Plug in z to verify it's a solution.</em>
- Substitute: -4 - (-10 - 6) = 12
- Subtract: -4 - (-16) = 12
- Simplify: -4 + 16 = 12
- Add: 12 = 12
Here we see that 12 does indeed equal 12.
∴ z = -10 is a solution of the equation.