Answer:
(a) 2996 units
(b) 1897 units
Step-by-step explanation:
p = 10,000(1 - 3/3 + e^-0.001x)
(a) p = $500
500 = 10,000(1 - 1 + e^-0.001x)
500/10,000 = e^-0.001x
e^-0.001x = 0.05
-0.001x = In 0.05 = -2.996
x = -2.996/-0.001 = 2996 units
(b) p = $1500
1500 = 10,000(1 - 3/3 + e^-0.001x)
1500/10,000 = (1 - 1 + e^-0.001x)
0.15 = e^-0.001x
-0.001x = In 0.15 = -1.897
x = -1.897/-0.001 = 1897 units
Answer:
168.168 square inches
Step-by-step explanation:
Given data
From the question, we can deduce that the side of the square is 28in
Hence the area of the square is
Area= 28^2
Area= 784 square inches
Also, the diameter of the circle is 28 in
The radius = 14 in
Area of circle= πr^2
Area= 3.142*14^2
Area= 3.142* 196
Area= 615.832 square inches
Therefore the area of the shaded region is
= Area of Square - Area of Circle
= 784-615.832
=168.168 square inches
Answer:
The ratio of the perimeter of ABCD to JKLM is;
b. 3/5
Step-by-step explanation:
Given that ABCD ~ JKLM, where;
DC and ML are corresponding sides
AD and JM are corresponding sides
AB and JK are corresponding sides
Therefore;
DC/ML = AD/JM = AB/JK
Plugging in the values of the variables gives;
9/15 = z/10 = x/12
∴ z = 10×9/15 = 6
x = 12×9/15 = 7.2
9/15 = 6/y
y = 6 ×15/9 = 10
The perimeter of the quadrilateral ABCD = AB + BC + DC + AD
∴ The perimeter of the quadrilateral ABCD = 6 + 6 + 9 + 7.2 = 28.2
The perimeter of the quadrilateral JKLM = JK + JM + ML + LK
∴ The perimeter of the quadrilateral JKLM = 12 + 10 + 15 + 10 = 47
The ratio of the perimeter of ABCD to JKLM = 28.2/47 =0.6 = 3/5.
We know that
[area of circular cookie]=pi*r²
r=2.5 in
[area of circular cookie]=pi*2.5²----> 19.63 in²
the answer is
the area of circular cookie is 19.63 in²
Answer:
B. 6
Step-by-step explanation:
The picture of the question in the attached figure
we know that
A translation and a reflection are rigid transformations, that produce congruent figures'
Remember that
Two figures are congruent if their corresponding sides and their corresponding angles are congruent
In this problem
Triangle JKL and Triangle J'K'L' are congruent
so
substitute the given values
solve for x
