The first 202020 licenses cost full price, which is \$41$41dollar sign, 41 each. Thus, the first 202020 licenses cost a total of \$41 \cdot 20 = \$820$41⋅20=$820dollar sign, 41, dot, 20, equals, dollar sign, 820.

Hint #2

The remaining licenses have a 10\%10%10, percent discount. That means they cost 90\%90%90, percent of the full price. Let's write 90\%90%90, percent in its decimal form, 0.90.90, point, 9, in order to multiply.

\qquad 0.9 \cdot \$41 = \$36.900.9⋅$41=$36.900, point, 9, dot, dollar sign, 41, equals, dollar sign, 36, point, 90

Hint #3

There are 40 - 20 = 2040−20=2040, minus, 20, equals, 20 licenses at the reduced price of \$36.90$36.90dollar sign, 36, point, 90. Thus, the remaining 202020 licenses cost a total of \$36.90 \cdot 20 = \$738$36.90⋅20=$738dollar sign, 36, point, 90, dot, 20, equals, dollar sign, 738.

Hint #4

To find the cost for all 404040 licenses, let's add the costs of the full price and discounted licenses.

\qquad \$820 + \$738 = \$1558$820+$738=$1558dollar sign, 820, plus, dollar sign, 738, equals, dollar sign, 1558

Hint #5

The business would pay \$1558$1558dollar sign, 1558 to buy the licenses.

3 0

3 years ago

The eccentricity e of an ellipse is defined as the number c/a, where a is the distance of a vertex from the center and c is the

Anna71 [15]

Answer:

Check below, please.

Step-by-step explanation:

Hi, there!

Since we can describe eccentricity as $e=\frac{c}{a}$

a) Eccentricity close to 0

An ellipsis with eccentricity whose value is 0, is in fact, a degenerate one almost a circle. An ellipse whose value is close to zero is almost a degenerate circle. The closer the eccentricity comes to zero, the more rounded gets the ellipse just like a circle. (Check picture, please)

$\frac{x^2}{a^2} +\frac{y^2}{b^2} =1 \:(Ellipse \:formula)\\a^2=b^2+c^2 \: (Pythagorean\: Theorem)\:a=longer \:axis.\:b=shorter \:axis)\\a^2=b^2+(0)^2 \:(c\:is \:the\: distance \: the\: Foci)\\\\a^2=b^2 \\a=b\: (the \:halves \:of \:each\:axes \:measure \:the \:same)$

b) Eccentricity =5

$5=\frac{c}{a} \:c=5a$

An eccentricity equal to 5 implies that the distance between the Foci has to be five (5) times larger than the half of its longer axis! In this case, there can't be an ellipse since the eccentricity must be between 0 and 1 in other words:

$If\:e=\frac{c}{a} \:then\:c>0 , and\: c>0 \:then \:1>e>0$

c) Eccentricity close to 1

In this case, the eccentricity close or equal to 1 We must conceive an ellipse whose measure for the half of the longer axis a and the distance between the Foci 'c' they both have the same size.

$a=c\\\\a^2=b^2+c^2\:(In \:the\:Pythagorean\:Theorem\: we \:should\:conceive \:b=0)$

$Then:\\\\a=c\\e=\frac{c}{a}\therefore e=1$

7 0

3 years ago