Answer:
Height of the box = 11.5 in
Step-by-step explanation:
Let h be the height of the box.
Assuming the volume of the Box is
.
Given:
Length = Height - 4 = h - 4
Width = 3 in
We need to find the height of the box.
Solution:
We know that the volume of the box.

Substitute all given value in above formula.

Rewrite the equation as:



whole equation divided by 3.

Use quadratic formula with

Put these a, b and c value in above equation.




For positive sign
h = 11.5 in
For negative sign

h = -7.5
We take positive value of h.
Therefore, the height of the box h = 11.5 in
If the half-life is t, then every t days, the amount of the radioactive isotope will be cut in half.
(1/2)^(number of half-lives) = 3%
number of half-lives = ln(0.03) / ln(0.5)
This gives the number of half-lives as 5.06.
Then 300 days = (5.06)(length of 1 half-life)
length of 1 half-life = 300 / 5.06 = 59.29 days
a is the beggining number and n is the amount of times needed to do this. if is 17 you would need to subtract 7 from the originol number 12 (a) that many times.
Answer:
<u>Triangle ABC and triangle MNO</u> are congruent. A <u>Rotation</u> is a single rigid transformation that maps the two congruent triangles.
Step-by-step explanation:
ΔABC has vertices at A(12, 8), B(4,8), and C(4, 14).
- length of AB = √[(12-4)² + (8-8)²] = 8
- length of AC = √[(12-4)² + (8-14)²] = 10
- length of CB = √[(4-4)² + (8-14)²] = 6
ΔMNO has vertices at M(4, 16), N(4,8), and O(-2,8).
- length of MN = √[(4-4)² + (16-8)²] = 8
- length of MO = √[(4+2)² + (16-8)²] = 10
- length of NO = √[(4+2)² + (8-8)²] = 6
Therefore:
and ΔABC ≅ ΔMNO by SSS postulate.
In the picture attached, both triangles are shown. It can be seen that counterclockwise rotation of ΔABC around vertex B would map ΔABC into the ΔMNO.
Answer:
7 - x meters
Step-by-step explanation:
This is a difference of squares. The formula for difference of squares is
a^2-b^2 = (a-b)(a+b)
So plug the numbers we know into this formula.
Area = lw
49 - x^2 = (a-b)(7+x)
49 - x^2 = (7-x)(7+x)
<u>The length is 7 - x meters.</u>