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3 years ago

Find two positive numbers a and b (witha≤b) whose sum is 88 and whose product is maximized.

1 answer:

3 0

a + b = 88 

ab = y

a = 88 - b

y = (88 - b)*b
 y = -b^2 + 88b

Take the derivative and set equal to 0

y' = -2b + 88 = 0

2b = 88
 b = 44

a=44

both numbers are 44

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Divide: 4/15÷8/25. What is the answer?

dybincka [34]

4/15 = 0.267
8/25 = 0.32
0.267/0.32 = 0.834375

3 0

2 years ago

The price of grapes at the grocery store is $1.44 per pound. There are approximately 0.45 kilograms in a pound. To the nearest c

Arlecino [84]

1 lb / 0.45 = 2.22

so approximately 2.22 kilograms per pound

divide price per pound by kilometers per pound

1.44 / 2.22 = 0.648

round to 0.65, so it costs $0.65 per kilogram

3 0

3 years ago

Scott buys candy that costs $7 per kilogram. He will spend at least $35 on candy. What are the possible numbers of kilograms he

slega [8]

Answer:

p ≥ 5

Scott will buy at least 5 kilograms of candy.

Step-by-step explanation:

Inequalities

The candy Scott buys cost $7 per kilogram.

Let's set p=number of kilograms of candy Scott will buy.

The money spent to buy p kilograms of candy is 7p dollars.

The condition states he will spend at least $35 on candies, thus the following inequality is formed:

7p ≥ 35

Dividing by 7:

p ≥ 35/7

Operating:

p ≥ 5

Scott will buy at least 5 kilograms of candy.

3 0

3 years ago

A teacher places n seats to form the back row of a classroom layout. Each successive row contains two fewer seats than the prece

Alex_Xolod [135]

Answer:

The number of seat when n is odd $S_n=\frac{n^2+2n+1}{4}$

The number of seat when n is even $S_n=\frac{n^2+2n}{4}$

Step-by-step explanation:

Given that, each successive row contains two fewer seats than the preceding row.

Formula:

The sum n terms of an A.P series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$=\frac{n}{2}[a+l]$

a = first term of the series.

d= common difference.

n= number of term

l= last term

$n^{th}$ term of a A.P series is

$T_n=a+(n-1)d$

n is odd:

n,n-2,n-4,........,5,3,1

Or we can write 1,3,5,.....,n-4,n-2,n

Here a= 1 and d = second term- first term = 3-1=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=1 and d=2

$n=1+(t-1)2$

⇒(t-1)2=n-1

$\Rightarrow t-1=\frac{n-1}{2}$

$\Rightarrow t = \frac{n-1}{2}+1$

$\Rightarrow t = \frac{n-1+2}{2}$

$\Rightarrow t = \frac{n+1}{2}$

Last term l= n,, the number of term $=\frac{ n+1}2$ , First term = 1

Total number of seat

$S_n=\frac{\frac{n+1}{2}}{2}[1+n}]$

$=\frac{{n+1}}{4}[1+n}]$

$=\frac{(1+n)^2}{4}$

$=\frac{n^2+2n+1}{4}$

n is even:

n,n-2,n-4,.......,4,2

Or we can write

2,4,.......,n-4,n-2,n

Here a= 2 and d = second term- first term = 4-2=2

Let $t^{th}$ of the series is n.

$T_n=a+(n-1)d$

Here $T_n=n$ , n=t, a=2 and d=2

$n=2+(t-1)2$

⇒(t-1)2=n-2

$\Rightarrow t-1=\frac{n-2}{2}$

$\Rightarrow t = \frac{n-2}{2}+1$

$\Rightarrow t = \frac{n-2+2}{2}$

$\Rightarrow t = \frac{n}{2}$

Last term l= n, the number of term $=\frac n2$ , First term = 2

Total number of seat

$S_n=\frac{\frac{n}{2}}{2}[2+n}]$

$=\frac{{n}}{4}[2+n}]$

$=\frac{n(2+n)}{4}$

$=\frac{n^2+2n}{4}$

4 0

2 years ago

Idk what the properties are I need help

slega [8]

Associative,commutative and distributive

3 0

2 years ago