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2 years ago

BRAINLIESt :P quick fast pls

Mathematics

2 answers:

Masteriza [31]2 years ago

4 0

Answer:

D. 2 1/6

Step-by-step explanation:

To visualize the problem easier, start by converting all the fractions to improper fractions, this will allow you to just add the denominators and simplify the question:

1/6, 5/6, 1 3/6 —> 1/6, 5/6, 9/6

Now that the sequence is rewritten, you can simply add 4 to the numerator, which equals 13/6.

Rewrite this again by finding how many times 6 goes into 13, and use the remainder as the numerator of the new fraction, which will end up as 2 1/6.

Molodets [167]2 years ago

4 0

The answer to this question is D.

You might be interested in

How do you tell if a number is irrational or rational?

Korvikt [17]

A rational number is any number that can be expressed as a fraction
(ex: -5/2, 4 1/7, and 8/13)

An irrational number is any number that cannot be expressed as a fraction; a non-repeating, non-terminating decimal (ex: pi, 0.121231234...)

Hope this helps! :)

6 0

3 years ago

Suppose a room is 5.2 m long by 4.3m wide and 2.9 m high and has an air conditioner that exchanges air at a rate of 1200 L/min.

Readme [11.4K]

Answer:

54 minutes

Step-by-step explanation:

From the question, we are given;

A room with dimensions 5.2 m by 4.3 m by 2.9 m
The exchange air rate is 1200 L/min

We are required to determine the time taken to exchange the air in the room;

First we are going to determine the volume of the room;

Volume of the room = length × width × height

= 5.2 m × 4.3 m × 2.9 m

= 64.844 m³

Then we should know, that 1 m³ = 1000 L

Therefore, we can convert the volume of the room into L

= 64.844 m³ × 1000 L

= 64,844 L

But, the rate is 1200 L/min

Thus, time = Volume ÷ rate

= 64,844 L ÷ 1200 L/min

= 54.0367 minutes

= 54 minutes

Therefore, it would take approximately 54 minutes

4 0

2 years ago

Complete the assignment on a separate sheet of paper Please attach pictures of your work.

Irina18 [472]

Answer:

TO FIND :-

Length of all missing sides.

FORMULAES TO KNOW BEFORE SOLVING :-

$\sin \theta = \frac{Side \: opposite \: to \: \theta}{Hypotenuse}$
$\cos \theta = \frac{Side \: adjacent \: to \: \theta}{Hypotenuse}$
$\tan \theta = \frac{Side \: opposite \: to \: \theta}{Side \: adjacent \: to \: \theta}$

SOLUTION :-

1) θ = 16°

Length of side opposite to θ = 7

Hypotenuse = x

$=> \sin 16 = \frac{7}{x}$

$=> \frac{7}{x} = 0.27563......$

$=> x = \frac{7}{0.27563....} = 25.39568.....$ ≈ 25.3

2) θ = 29°

Length of side opposite to θ = 6

Hypotenuse = x

$=> \sin 29 = \frac{6}{x}$

$=> \frac{6}{x} = 0.48480......$

$=> x = \frac{6}{0.48480....} = 12.37599.....$ ≈ 12.3

3) θ = 30°

Length of side opposite to θ = x

Hypotenuse = 11

$=> \sin 30 = \frac{x}{11}$

$=> \frac{x}{11} = 0.5$

$=> x = 0.5 \times 11 = 5.5$

4) θ = 43°

Length of side adjacent to θ = x

Hypotenuse = 12

$=> \cos 43 = \frac{x}{12}$

$=> \frac{x}{12} = 0.73135......$

$=> x = 12 \times 0.73135.... = 8.77624....$ ≈ 8.8

5) θ = 55°

Length of side adjacent to θ = x

Hypotenuse = 6

$=> \cos 55 = \frac{x}{6}$

$=> \frac{x}{6} = 0.57357......$

$=> x = 6 \times 0.57357.... = 3.44145....$ ≈ 3.4

6) θ = 73°

Length of side adjacent to θ = 8

Hypotenuse = x

$=> \cos 73 = \frac{8}{x}$

$=> \frac{8}{x} = 0.29237......$

$=> x = \frac{8}{0.29237.....} = 27.36242.....$ ≈ 27.3

7) θ = 69°

Length of side opposite to θ = 12

Length of side adjacent to θ = x

$=> \tan 69 = \frac{12}{x}$

$=> \frac{12}{x} = 2.60508......$

$=> x = \frac{12}{2.60508....} = 4.60636....$ ≈ 4.6

8) θ = 20°

Length of side opposite to θ = 11

Length of side adjacent to θ = x

$=> \tan 20 = \frac{11}{x}$

$=> \frac{11}{x} = 0.36397......$

$=> x = \frac{11}{0.36397....} =30.22225....$ ≈ 30.2

5 0

2 years ago

an online seed supplier packages a seed mix that costs the company 20.70$ per pound the mix includes poppy seeds costing 24$ per

seraphim [82]

Answer:

The quantity of clover seeds in pounds to be added is 11.14

Step-by-step explanation:

Let

x -----> the quantity of poppy seeds in pounds

y -----> the quantity of clover seeds in pounds

we know that

$24(x)+13(y)=20.70(x+y)$ -----> equation A

$x=26\ pounds$ ----> equation B

substitute equation B in equation A and solve for y

$24(26)+13(y)=20.70(26+y)$

$624+13y=538.2+20,70y$

$20.70y-13y=624-538.2$

$7.7y=85.8$

$y=11.14\ pounds$

therefore

The quantity of clover seeds in pounds to be added is 11.14

8 0

3 years ago

Read 2 more answers

There are eight different jobs in a printer queue. Each job has a distinct tag which is a string of three upper case letters. Th

N76 [4]

Answer:

a. 40320 ways

b. 10080 ways

c. 25200 ways

d. 10080 ways

e. 10080 ways

Step-by-step explanation:

There are 8 different jobs in a printer queue.

a. They can be arranged in the queue in 8! ways.

No. of ways to arrange the 8 jobs = 8!

= 8*7*6*5*4*3*2*1

No. of ways to arrange the 8 jobs = 40320 ways

b. USU comes immediately before CDP. This means that these two jobs must be one after the other. They can be arranged in 2! ways. Consider both of them as one unit. The remaining 6 together with both these jobs can be arranged in 7! ways. So,

No. of ways to arrange the 8 jobs if USU comes immediately before CDP

= 2! * 7!

= 2*1 * 7*6*5*4*3*2*1

= 10080 ways

c. First consider a gap of 1 space between the two jobs USU and CDP. One case can be that USU comes at the first place and CDP at the third place. The remaining 6 jobs can be arranged in 6! ways. Another case can be when USU comes at the second place and CDP at the fourth. This will go on until CDP is at the last place. So, we will have 5 such cases.

The no. of ways USU and CDP can be arranged with a gap of one space is:

6! * 6 = 4320

Then, with a gap of two spaces, USU can come at the first place and CDP at the fourth. This will go on until CDP is at the last place and USU at the sixth. So there will be 5 cases. No. of ways the rest of the jobs can be arranged is 6! and the total no. of ways in which USU and CDP can be arranged with a space of two is: 5 * 6! = 3600

Then, with a gap of three spaces, USU will come at the first place and CDP at the fifth. We will have four such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 4 * 6!

Then, with a gap of four spaces, USU will come at the first place and CDP at the sixth. We will have three such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 3 * 6!

Then, with a gap of five spaces, USU will come at the first place and CDP at the seventh. We will have two such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 2 * 6!

Finally, with a gap of 6 spaces, USU at first place and CDP at the last, we can arrange the rest of the jobs in 6! ways.

So, total no. of different ways to arrange the jobs such that USU comes before CDP = 10080 + 6*6! + 5*6! + 4*6! + 3*6! + 2*6! + 1*6!

= 10080 + 4320 + 3600 + 2880 + 2160 + 1440 + 720

= 25200 ways

d. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways. Similarly, if LPW comes last, the remaining 7 jobs can be arranged in 7! ways. so, total no. of different ways in which the eight jobs can be arranged is 7! + 7! = 10080 ways

e. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways in the queue. Similarly, if QKJ comes second-to-last then also the jobs can be arranged in the queue in 7! ways. So, total no. of ways to arrange the jobs in the queue is 7! + 7! = 10080 ways

3 0

3 years ago