Reduce a 24 cm by 36 cm photo to 3/4 original size.
The most logical way to do this is to keep the width-to-height ratio the same: It is 24/36, or 2/3. The original photo has an area of (24 cm)(36 cm) = 864 cm^2.
Let's reduce that to 3/4 size: Mult. 864 cm^2 by (3/4). Result: 648 cm^2.
We need to find new L and new W such that W/L = 2/3 and WL = 648 cm^2.
From the first equation we get W = 2L/3. Thus, WL = 648 cm^2 = (2L/3)(L).
Solve this last equation for L^2, and then for L:
2L^2/3 = 648, or (2/3)L^2 = 648. Thus, L^2 = (3/2)(648 cm^2) = 972 cm^2.
Taking the sqrt of both sides, L = + 31.18 cm. Then W must be 2/3 of that, or W = 20.78 cm.
Check: is LW = (3/4) of the original 864 cm^2? YES.
Answer:
(0, 4.5)
Step-by-step explanation:
*The equation can be put into Desmos, to find the point, but the work to prove it is here*
f(x)=c/1+Ae^-Bx
Y=C
C=18 A=3 -B=-0.1
*Replace x with 0 in the equation, so you know 0 is the x value, and it leads you to the y value*
f(0)=18/1+3e^-o.1(0)
= 18/1+3e^0
=18/1+3(1)
=18/1+3
=18/4
=4.5
x=0 y=4.5
Maximum growth rate = (x,y) --> (0, 4.5)
Hope this helps:))!!
Answer:
See attached
Step-by-step explanation:
Given function:
Table and graph are attached
Zeros are included in the graph
<u>Zero's are obtained:</u>
x = 0 ⇒ y = 8
y = 0 ⇒ Solving quadratic equation
- -2x² + 5x + 8 = 0
- x = (-5 ± √(25 + 2*4*8))/-4
- x = 3.608
- x = -1.108
So zeros are (0, 8), (3.608, 0) and (-1.108, 0)
First, you have to set a system of equations to determine the number of fiction and of nonfiction books.Call f the number of fiction books and n the number of nonfiction books. Then 400 = f + n. And f = n + 40 => n = f - 40 => 400 = f + f - 40 => 400 - 40 = 2f => f = 360 / 2 = 180. Now to find the probability of picking two fiction books, take into account the the Audrey will pick from 180 fiction books out of 400, and Ryan will pick from 179 fiction books out of 399, so the probability will be<span> (180/ 400) * (179/399) = 0.20 (rounded to two decimals). Answer: 0.20</span>
A polynomial function is a function involving only non-negative (positive and 0) integer powers of 'x', i.e. all terms with ' 'x' have a non-negative integer power, such as a quadratic
and so on.
Option E is the answer.
Rest all are wrong answers. they do not have positive integers.