Answer:
Maximum volume error = ±540 cm³
Relative error = 0.02
Percentage error = 2%
Step-by-step explanation:
Relative error : The ratio of volume error to the total volume.
Percentage error: The product of relative error and 100.
The volume of a cube is = 
v =x³
Differentiate with respect to x
Here are x = 30 cm and dx= ±0.2 cm
∴ dv = 3×(30 cm)² (±0.2 cm)
=±540 cm³
The volume of the cube = 30³ cm ³ = 27,000 cm³
Then the relative error
= 0.02.
The percentage error
= (0.02×100)
=2%
So, she has 550 dollars saved. If the junior season pass costs 315, you subtract that from 550. You should then have 235 dollars. Next, you divide 235 by 45 dollars per lesson, and you should get 5.22. You can't buy half a lesson, so she only has enough for 5 lessons.
Okay so in school I am doing this litterly right now. So all u have to do is when it is a different sign you subtract, when it is the same you add. So think of it as, Different/Difference and Same/Sum. when the sign is different u find the difference and when u have the same sign you find the sum.
Example 1: -3+8= 5
Use the same sign as the largest number
Example 2: 5+(-10)= -5
Hopefully This helped! It is easier then you may think once you get the hang of it! ( You get the hang of it REALLY fast ) :)
Answer:
(5 t ) cubed = 5 cubed . t cubed = 125 t cubed applies the power of a product rule to simplify (5 t) cubed ⇒ 3rd answer
Step-by-step explanation:
Let us revise some rules of exponents
× 
= 
×÷ = 
= 
= . 
To simplify 
∵ 5t means 5 × t
∵ Both of them are cubed
- Use the 4th rule above
∴ = 
∵ (5)³ = 5 × 5 × 5 = 125
∴ = = 125 t³
(5 t ) cubed = 5 cubed . t cubed = 125 t cubed applies the power of a product rule to simplify (5 t) cubed
Answer:
The ship is located at (3,5)
Explanation:
In the first test, the equation of the position was:
5x² - y² = 20 ...........> equation I
In the second test, the equation of the position was:
y² - 2x² = 7 ..............> equation II
This equation can be rewritten as:
y² = 2x² + 7 ............> equation III
Since the ship did not move in the duration between the two tests, therefore, the position of the ship is the same in the two tests which means that:
equation I = equation II
To get the position of the ship, we will simply need to solve equation I and equation II simultaneously and get their solution.
Substitute with equation III in equation I to solve for x as follows:
5x²-y² = 20
5x² - (2x²+7) = 20
5x² - 2y² - 7 = 20
3x² = 27
x² = 9
x = <span>± </span>√9
We are given that the ship lies in the first quadrant. This means that both its x and y coordinates are positive. This means that:
x = √9 = 3
Substitute with x in equation III to get y as follows:
y² = 2x² + 7
y² = 2(3)² + 7
y = 18 + 7
y = 25
y = +√25
y = 5
Based on the above, the position of the ship is (3,5).
Hope this helps :)
