We will define the following variables to solve the problem:
s: normal speed
t: normal time
We write the following system of equations:
6 = s * t
6 = (s + 3) * (t-7)
Rewriting the equation two we have:
6 = s * t -7s + 3t -21
Substituting:
s = 6 / t
6 = (6 / t) * t -7 (6 / t) + 3t -21
Multiplying both sides by t:
6t = 6t - 42 + 3t ^ 2 - 21t
Rewriting:
3t ^ 2 - 21t - 42 = 0
We take the positive root:
t = 7/2 + root (105) / 2
t = 8.62 h
We look for the value of the speed:
s = 6 / t
s = 6 / 8.62
s = 0.696 km / h
Answer:
she did travel 0.696 kilometers per hour
Answer:
2 1/4 yards.
Step-by-step explanation:
Width = area / length
Width = 3/4 / 1/3
= 3/4 * 3
= 9/4
= 2 1/4 yards.
The width is actually the length!! but nvm.
Answer:
3/13 or 23.1%
Step-by-step explanation:
Answer:
<em>Similar: First two shapes only</em>
Step-by-step explanation:
<u>Triangle Similarity Theorems
</u>
There are three triangle similarity theorems that specify under which conditions triangles are similar:
If two of the angles are congruent, the third angle is also congruent and the triangles are similar (AA theorem).
If the three sides are in the same proportion, the triangles are similar (SSS theorem).
If two sides are in the same proportion and the included angle is equal, the triangles are similar (SAS theorem).
The first pair of shapes are triangles that are both equilateral and therefore have all of its interior angles of 60°. The AAA theorem is valid and the triangles are similar.
The second pair of shapes are parallelograms. The lengths are in the proportion 6/4=1,5 and the widths are in proportion 3/2=1.5, thus the shapes are also similar.
The third pair of shapes are triangles whose interior acute angles are not congruent. These triangles are not similar
Answer:
299.99 miles
Step-by-step explanation:
Since the plane traveled due west,
The total angle is 49.17 + 90
Represent that with θ
θ = 49.17 + 90
θ = 139.17.
Represent the sides as
A = 170
B = 150
C = unknown
Since, θ is opposite side C, side C can be calculated using cosine formula as;
C² = A² + B² - 2ABCosθ
Substitute values for A, B and θ
C² = 150² + 170² - 2 * 150 * 170 * Cos 139.17
C² = 22500 + 28900 - 51000 * Cos 139.17
C² = 51400 - 51000 (−0.7567)
C² = 51400 + 38,591.7
C² = 89,991.7
Take Square Root of both sides
C = 299.9861663477167
C = 299.99 miles (Approximated)
Hence, the distance between the plane and the airport is 299.99 miles
