Simplify the radical expression -21√27x^9

Two canoes travel down a river, starting at 9:00 a.M. One canoe travels twice as fast as the other. After 4.5 hours, the canoes

Vedmedyk [2.9K]

Answer:

$v_1=3.33 \ miles/hour\\v_2=6.66 \ miles/hour$

Step-by-step explanation:

Constant Speed Motion

When the object travels the same distances in the same period of time, the speed is constant and is can be calculated as

$\displaystyle v=\frac{x}{t}$

where x is the distance and t is the time

There are two canoes traveling at speeds v1 and v2. One of them has twice the speed as the other. Let's say

$v_2=2v_1$

When t hours have passed, the canoes have traveled x1 and x2 respectively.

$x_1=v_1t$

$x_2=v_2t$

Since

$v_2=2v_1$

Then

$x_2=2v_1t$

The distance between them is

$X=x_2-x_1=2v_1t-v_1t$

$X=v_1t$

That distance is known and equals 11.25 miles, thus

$v_1t=15$

Solving for v1 and using t=4.5 hours

$\displaystyle v_1=\frac{15}{4.5}$

$v_1=3.33 \ miles/hour$

$v_2=2\times 3.33$

$\boxed{v_2=6.66 \ miles/hour}$

6 0

2 years ago

X+3y=14 y=3x-14 Solving systems using substitution How do i do this?

Free_Kalibri [48]

X + 3y = 14
x + 3(3x - 14) = 14
x + 9x - 42 = 14
10x - 42 = 14
 + 42 + 42
 10x = 56
 10 10
 x = 5.6
5.6 + 3y = 14
- 5.6 - 5.6
 3y = 8.4
 3 3
 y = 2.7
 (x, y) = (2.7, 5.6)

6 0

3 years ago

Read 2 more answers

In the diagram AB =AD and

Greeley [361]

Answer:

AC ≅ AE

Step-by-step explanation:

According to the SAS Congruence Theorem, for two triangles to be considered equal or congruent, they both must have 2 corresponding sides that are of equal length, and 1 included corresponding angle that is of the same measure in both triangles.

Given that in ∆ABC and ∆ADE, AB ≅ AD, and <BAC ≅ DAE, the additional information we need to prove that ∆ABC ≅ ADE is AC ≅ AE. This will satisfy the SAS Congruence Theorem. As there would be 2 corresponding sides that are congruent, and 1 corresponding angle in both triangles that are congruent to each other.

8 0

3 years ago

Michael’s weight can be represented by the expression 72x^5. Al’s weight can be represented by the expression 9x^7.

podryga [215]

Answer:

a) $\frac{72x^5}{9x^7}$