Units are squared when dealing with area but not when dealing with perimeter

The original price of the phone is $137.99. It is now on sale for 1/10 off the original price. This means that the discount on the original price is 1/10 × 137.99 = 13.799. The new price will be the original price - the discount.

The new price is 137.99 - 13.799 = $124.191

Tom has a coupon for an extra 5% off the sale price. It means that Tom would pay 5% off $124.191. The amount that Tom would pay will be

124.191 - (5/100 × 124.191) = 124.191 - 6.20955 = $117.98

The difference between the original price and the price that Tom will pay is

137.99 - 117.98 = $20.01

7 0

3 years ago

Consider a rectangle of length L inches and width W inches. Find a formula for the perimeter of the rectangle. Use upper case le

malfutka [58]

Answer:

a) P=2(L+W)

b) $\frac{dp}{dt}=2\frac{dL}{dt}+2\frac{dW}{dt}$

c)-2 inch/hour

Step-by-step explanation:

given:

length of the rectangle as L inches

width of the rectangle as W inches

a) The perimeter is defined as the measure of the exterior boundaries

therefore, for the rectangle the perimeter 'P' will be

P= length of AB+BC+CD+DA (A,B,C and D are marked on the figure attached)

Now from figure

P= L+W+L+W

OR

=> P=2L+2W .....................(1)

b)now dp/dt can be found as by differentiating the equation (1)

$\frac{dP}{dt}=2(\frac{dL}{dt} )+2(\frac{dW}{dt} )$ .............(2)

c)Now it is given for the part c of the question that

L=40 inches

W=104 inches

dL/dt=2 inches/hour

dW/dt= -3 inches/hour (here the negative sign depicts the decrease in the dimension)

substituting the above values in the equation (2) we get

$\frac{dP}{dt}=2(2)+2(-3)$

$\frac{dP}{dt}=4-6=-2 inches/hour$

5 0

3 years ago

use the definition of countinuity to find the value of k so that the function is continuous for all real numbers

liubo4ka [24]

First of all, recall the definition of absolute value:

$|x| = \begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

So if x < 4, then x - 4 < 0, so |x - 4| = -(x - 4), and the first case in h(x) reduces to

$\dfrac{|x-4|}{x-4}=\dfrac{-(x-4)}{x-4} = -1$

Next, in order for h(x) to be continuous at x = 4, the limits from either side of x = 4 must be equal and have the same value as h(x) at x = 4. From the given definition of h(x), we have

$h(4) = 5k-4\cdot4 = 5k-16$

Compute the one-sided limits:

• From the left:

$\displaystyle \lim_{x\to4^-}h(x) = \lim_{x\to4}\frac{|x-4|}{x-4} = \lim_{x\to4}(-1) = -1$

• From the right:

$\displaystyle \lim_{x\to4^+}h(x) = \lim_{x\to4}(5k-4x) = 5k-16$

If the limits are to be equal, then

-1 = 5k - 16

Solve for k :

-1 = 5k - 16

15 = 5k

k = 3

3 0

3 years ago