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attashe74 [19]
2 years ago
9

Mai spends 7 and 3/5 hours in school each day. Her lunch period is 30 minutes long, and she spends a total of 42 minutes switchi

ng rooms between classes. The rest of her day is spent in 6 classes that are all the same length. How long is each class?
Mathematics
1 answer:
Sonja [21]2 years ago
8 0
First, we solve for the number of minutes in 7 and 3/5 hours by multiplying the number by 60 giving us,
                            (7 + 3/5) x (60) = 456 minutes
Spending 30 minutes for lunch will leave her with 426 minutes. Then, spending 42 minutes for switching of classes will finally give her 414 minutes. 

We then divide this value by 6 (for her 6 classes) giving us 69 minutes. Thus, each class is 69 minutes long. 
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PLEASE ANSWER ASAP WILL MARK BRAINLIEST
katrin2010 [14]

Answer:

48 hats and 104 shirts

Step-by-step explanation:

These are the equations you build from the problem:

h + s = 152

8.50h + 12s = 1656

This is how I solved them:

s= 152-h

8.5h + 12(152-h) = 1656

8.5h + 1824 - 12h = 1656

Solve for h

h= 48

Put this into first equation (h +s = 152) to get s

8 0
2 years ago
Read 2 more answers
Consider the following problem: A farmer with 950 ft of fencing wants to enclose a rectangular area and then divide it into four
Alina [70]

Answer:

Step-by-step explanation:

(a)

Suppose we came up with an ideology whereby we pick a value for the length including the length dividing the inside into 4 parts(5 parallel sides), then we can get the value for breath by using the following process.

Let assume the length of the rectangle is 50;

Then, the breath can be calculated as follows:

= 50 × 5 = 250   ( since the breath is divided into 5 parallel sides)

The fencing is said to be 950 ft

So, 950 - 250 = 700

Then divided by 2, we get:

= 700/2

= 350

So for the first diagram; the length = 50 and the breath = 350

The area = 50 × 350 = 17500 ft²

Now, let's go up a little bit.

If the length increase to 100;

Then 100 × 5 = 500

⇒ 950 - 500 = 450

⇒ 450/2 = 225

The area = 225 × 100 = 22500 ft²

Suppose the length increases to 150

Then 150 × 5 = 750

⇒ 950 - 750 = 200

⇒ 200/2 = 100

The area = 150 × 100 = 15000 ft²

The diagrams for each of the outline above can be seen in the image attached below.

(b) The diagram illustrating the general solution can be seen in the second image provided below.

(c) The expression for  the total area A in terms of both x and y is:

Area A = x×y

(d) Recall that:

The fencing is said to be 950 ft.

And the length is divided inside into 5 parallel sides;

Then:

5x + 2y = 950  (from the illustration in the second image below)

2y  = 950 - 5x

y = \dfrac{950}{2} - \dfrac{5}{2}x

y = 475- \dfrac{5}{2}x

(e)

From (c); replace the value of y in (d) into (c)

Then:

Area A = x×y

f(x)= x\times ( 475 -\dfrac{5}{2}x)

Open brackets

f(x)= ( 475 x-\dfrac{5}{2}x^2)

(f)

By differentiating what we have in (e)

f(x)= ( 475 x-\dfrac{5}{2}x^2)

f'(x)= ( 475 (1)-\dfrac{5}{2}(2x))

f'(x)= 475 -5x

\implies  475 = 5x

x = 475/5

x = 95

From (d):

y = 475- \dfrac{5}{2}x

y = 475- \dfrac{5}{2}(95)

y =237.5

∴

Area A = x × y

Area A = 95 × 237.5

Area A = 22562.5 ft²

5 0
2 years ago
a dance club spent $922 on 40 items.The item include some hats and pairs of shoes.Each hat cost $19 and each pair of shoes cost
Sergio039 [100]

1. H+S=40

2. 19H+25S=922

From 1,

19H+19S=760

Subtract this from 2 to eliminate H,

19H+25S-19H-19S=922-760

6S=162

Solve for S, then use either equation to solve for H.

4 0
2 years ago
Paige works at a department store and has a commission rate of 3 percent. Her sales for the quarter were $4,596. If she earns $5
ANTONII [103]
How you would set this up would be as follows:

(50*30)+4,596*.03

The 50*30 is her earning $50 a day for 30 days. Then to find her commission earnings you multiply her sales of 4,596 times the commission rate of 3%(when you move the decimal place you get .03)

Add them together and you get her total amount earned of $1,637.88

Hope this helped!


6 0
2 years ago
Read 2 more answers
Demand for Tablet Computers The quantity demanded per month, x, of a certain make of tablet computer is related to the average u
soldier1979 [14.2K]

x = f ( p ) = \frac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } } \\\\ \qquad { p ( t ) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \quad ( 0 \leq t \leq 60 ) }

Answer:

12.0 tablet computers/month

Step-by-step explanation:

The average price of the tablet 25 months from now will be:

p ( 25) = \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { 25 } } + 200 \\= \dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \times 5 } + 200\\\\=\dfrac { 400 } { 1 + \dfrac { 5 } { 8 } } + 200\\p(25)=\dfrac { 5800 } {13}

Next, we determine the rate at which the quantity demanded changes with respect to time.

Using Chain Rule (and a calculator)

\dfrac{dx}{dt}= \dfrac{dx}{dp}\dfrac{dp}{dt}

\dfrac{dx}{dp}= \dfrac{d}{dp}\left[{ \dfrac { 100 } { 9 } \sqrt { 810,000 - p ^ { 2 } } }\right] =-\dfrac{100}{9}p(810,000-p^2)^{-1/2}

\dfrac{dp}{dt}=\dfrac{d}{dt}\left[\dfrac { 400 } { 1 + \dfrac { 1 } { 8 } \sqrt { t } } + 200 \right]=-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}

Therefore:

\dfrac{dx}{dt}= \left[-\dfrac{100}{9}p(810,000-p^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt { t } \right]^{-2}t^{-1/2}\right]

Recall that at t=25, p(25)=\dfrac { 5800 } {13} \approx 446.15

Therefore:

\dfrac{dx}{dt}(25)= \left[-\dfrac{100}{9}\times 446.15(810,000-446.15^2)^{-1/2}\right]\left[-25\left[1 + \dfrac { 1 } { 8 } \sqrt {25} \right]^{-2}25^{-1/2}\right]\\=12.009

The quantity demanded per month of the tablet computers will be changing at a rate of 12 tablet computers/month correct to 1 decimal place.

8 0
2 years ago
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