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3 years ago

A rectangle has an area of 99 square in. What is the length of the long side?

Mathematics

1 answer:

OLga [1]3 years ago

8 0

Answer:

The longer length is either 99, 33 or 11 inches

Step-by-step explanation:

Given

$Area = 99$

Required

The length of the longer side

Area of a rectangle is:

$Area = Length * Width$

List out all possible products that give 99 [Assume the dimension are whole numbers]

$Length * Width = 99 * 1 = 99$

$Length * Width = 33 * 3 = 99$

$Length * Width = 11 * 9 = 99$

List out the longer factors

$L = \{99, 33, 11\}$

<em>Hence, the longer length is either 99, 33 or 11 inches</em>

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Problem 3.2.14a<br><br> Show that 2^2x+1 +1 is divisible by 3.

elena-s [515]

Answer:

The given expression is divisible by 3 for all natural values of x.

Step-by-step explanation:

The given expression is

$2^{2x+1}+1$

For x=1,

$2^{2(1)+1}+1=2^{3}+18+1=9$

9 is divisible by 3. So, the given statement is true for x=1.

Assumed that the given statement is true for n=k.

$2^{2k+1}+1$

This expression is divisible by 3. So,

$2^{2k+1}+1=3n$ .... (1)

For x=k+1

$2^{2(k+1)+1}+1$

$2^{2k+2+1}+1$

$2^{(2k+1)+2}+1$

$2^{2k+1}2^2+1$

Using equation (1), we get

$(3n-1)2^2+1$

$(3n)2^2-2^2+1$

$(3n)2^2-4+1$

$(3n)4-3$

$3(4n-1)$

This expression is also divisible by 3.

Therefore the given expression is divisible by 3 for all natural values of x.

4 0

4 years ago

A jar of sweet contains 5 yellow sweets, 4 red sweets, 8 green sweets, 4 orange sweets and 3 white sweets. Ola chose a sweet at

Sveta_85 [38]

Answer:

9/24 or 37.5%

hope this helps :)

8 0

4 years ago

Read 2 more answers

At a school with 100? students, 33 were taking? Arabic, 32 ?Bulgarian, and 40 Chinese. 9 students take only? Arabic, 12 take onl

Stels [109]

Answer: The answer is 4 and 32.

Step-by-step explanation: Let "A", "B" and "C" represents the set of students who were taking Arabic, Bulgarian and Chinese respectively.

The, according to the given information, we have

$n(A)=33,~~n(B)=32,~~n(C)=40,~~n(A\cap B)=14.$

Let 'p' represents the number of students who take all the three languages, then

$n(A\cap B\cap C)=p.$

Also,

$n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)=9\\\\\Rightarrow n(A\cap B)+n(A\cap C)=24+p~~~~~~~~~~~~~~(a),\\\\n(A\cap B)+n(B\cap C)=20+p~~~~~~~~~~~~~~(b),\\\\n(B\cap C)+n(A\cap C)=20+p~~~~~~~~~~~~~~~(c).$

From here, we get after subtracting equation(c) from (b) that

$n(A\cap B)=n(A\cap C)=14.$

Therefore,

$p=14+14-24=4,$ and from equation (a), we find

$n(B\cap C)=24-14=10.$

Thus,

$n(A\cap B\cap C)=4$ and

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\\Rightarrow n(A\cap B\cap C)=33+32+40-9-12-20+4\\\\\Rightarrow n(A\cap B\cap C)=68.$