Answer:
Total trade discount on 30 boxes = $201.555
Step-by-step explanation:
1 box = $14.93
And a trade discount of 45% is allowed on a single box,
Therefore, the discount on one $14.93 box will be
45% × $14.93 = $6.7185
So, if there are now 30 boxes, the trade discount on all 30 boxes will be
30 × (trade discount on one box) = 30 × ($6.7185) = $201.555
Hope this helps!
Answer:
Step-by-step explanation:
9*21*2+18*9*2+21*18+35*20+(35*20-21*18)+35*15*2+15*20*2
= 378 + 324+ 378 + 700 + 700 - 378 + 1050 + 600
= 3752
<h2>Solving Equations with Absolute Values</h2><h3>
Answer:</h3>
and 
<h3>
Step-by-step explanation:</h3>
Solving for the Positive Absolute Value:
Solving for the Negative Absolute Value:
Answer:
<h3>(a) 3^9</h3>
Step-by-step explanation:
<h3> IN EXPONNETIAL FORM :</h3>
<h3>– (a^m× a^n) = (a^m+n)</h3>
<h3>3^4 × 3^5 = 3^4+5 = </h3>
<h2>3^9 </h2>
<h3 />
Answer:
(f⁻¹)'(b) = 1/f'(f⁻¹(b)) = 1/f'(a)
Step-by-step explanation:
The function f⁻¹(x) is the reflection of the function f(x) across the line y=x. Every point (a, b) that is on the graph of f(x) is reflected to be a point (b, a) on the graph of f⁻¹(x).
Any line with slope m reflected across the line y=x will have slope 1/m. (x and y are interchanged, so m=∆y/∆x becomes ∆x/∆y=1/m) Since f'(x) is the slope of the tangent line at (x, f(x)), 1/f'(x) will be the slope of the tangent line at (f(x), x).
Replacing x with f⁻¹(x) in the above relation, you get ...
... (f⁻¹)'(x) = 1/f'(f⁻¹(x)) will be the slope at (x, f⁻¹(x))
Putting your given values in this relation, you get
... (f⁻¹)'(b) = 1/f'(f⁻¹(b)) = 1/f'(a)
