Answer:
Step-by-step explanation:
Using the distributive property to expand, we have
Squaring each term, we have
The mistake was that 
not 
Using the fact that the right answer would be
.
Volume of the wrapper = pi * 0.7^2 * 10 = 4.9pi cm^2
Volume of one tablet = 0.7^2 * 0.5 pi = 0.245pi
Number of tablets in the wrappe = 4.9pi / 0.245pi = 20 (answer)
The required equation of line is y=-6x-57.
Step-by-step explanation:
The point slope form of a line is
y-y_1=m(x-x_1)y−y
1
=m(x−x
1
)
Where m is slope of the line.
It is given that a line that passes through (–9, –3) and has a slope of –6.
Substitute m=-6, x₁=-9 and y₁=-3 in the above equation.
y-(-3)=-6(x-(-9))y−(−3)=−6(x−(−9))
y+3=-6x-54y+3=−6x−54
Subtract 3 from both the sides.
y=-6x-54-3y=−6x−54−3
y=-6x-57y=−6x−57
Therefore the required equation of line is y=-6x-57.
Answer:
$247.50
Step-by-step explanation:
33/100x750=247.50
Since 33% in a fraction is 33/100, use that x 750 and you will get how much the computer gets marked down.
Hope this helps! Thanks.
Here is the correct format for the question
At 2:00 PM a car's speedometer reads 30 mi/h. At 2:15 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:15 the acceleration is exactly 80 mi/h².Let v(f) be the velocity of the car t hours after 2:00 PM.Then 
. By the Mean Value Theorem, there is a number c such that 0 < c < 
with v'(c) = . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly 80 mi/h^2.
Answer:
Step-by-step explanation:
From the information given :
At 2:00 PM ;
a car's speedometer v(0) = 30 mi/h
At 2:15 PM;
a car's speedometer v(1/4) = 50 mi/h
Given that:
v(f) should be the velocity of the car t hours after 2:00 PM
Then will be:
= 20 × 4/1
= 80 mi/h²
By the Mean value theorem; there is a number c such that :
with 
