Answer:
4z4−6z3−2z2−20z+1
Step-by-step explanation:
4z5−14z4+10z3−16z2+41z−2
z−2
=
4z5−14z4+10z3−16z2+41z−2
z−2
=
(z−2)(4z4−6z3−2z2−20z+1)
z−2
=4z4−6z3−2z2−20z+1
(theres no more like terms so u cant)
Answer:
m∠A = 106°
Step-by-step explanation:
All three sides of the triangle are given in the question.
To find the measure of included angle A between two sides we will apply cosine formula,
BC²= AB² + AC² - 2(AB)(AC)cosA
(13)² = (10)² + (6)² - 2(10)(6)cosA
169 = 100 + 36 - 120cosA
cosA = 
cosA = -0.275
A = 
A = 105.96°
A ≈ 106°
Therefore, m∠A = 106° will be the answer.
So a general harmonic equation is Asin(2πt/T). Little t is just a variable. 6 inches is the height of one maximum and one minimum from equilibrium, so that is the amplitude A. T is the period, or the time of one cycle (or wavelength), which is 2 seconds.
The equation is therefore (in terms of seconds and inches), 6*sin(2πt/2), or 6*sin(πt).
Overall dimensions of the page in order to maximize the printing area is page should be 11 inches wide and 10 inches long .
<u>Step-by-step explanation:</u>
We have , A page should have perimeter of 42 inches. The printing area within the page would be determined by top and bottom margins of 1 inch from each side, and the left and right margins of 1.5 inches from each side. let's assume width of the page be x inches and its length be y inches So,
Perimeter = 42 inches
⇒ 
width of printed area = x-3 & length of printed area = y-2:
area = 

Let's find
:
=
, for area to be maximum
= 0
⇒ 
And ,

∴ Overall dimensions of the page in order to maximize the printing area is page should be 11 inches wide and 10 inches long .