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3 years ago

Find the measures of the supplementary angles if their measures are in ratio of 2:3

Mathematics

2 answers:

egoroff_w [7]3 years ago

8 0

Answer:

Two supplementary angles are 72° and 108° .

Step-by-step explanation:

we have given the measure of ratio of two angles is 2:3.

Let the two angles be 2x and 3x.

We know the sum of supplementary angle is 180°

Therefore,

2x + 3x = 180°

or 5x = 180°

or x = 180°/5

or x =36°

Hence, the first angle = 2x =2×36 = 72°

And Second angle = 3x = 3×36 = 108°

pentagon [3]3 years ago

4 0

Answer:

The measures are 72 degrees and 108 degrees.

Step-by-step explanation:

Supplementary angles means that the two angles add up to 180 degrees. So if we let the smallest angle equal x, we can say

x+(3/2)x=180

5/2x=180

Multiplying both sides by 2/5, we find

x=180*2/5=72

Thus, the smallest angle is 72 degrees. Now we subtract 72 from 180 to get 108. Checking our work, we see that 72:108=2:3. So, the measures are 72 degrees and 108 degrees.

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seropon [69]

Answer: Its factored form will be

$6n(n-1)(n^2-3n-3)$

Step-by-step explanation:

Since we have given that

$6n^4-24n^3+18n$

So, we need to factor ,

$6n(n^3-4n^2+3)$

Now,

$\text{Let }p(n)=n^3-4n^2+3$

Now, by hit and trial method, we put n=1,

$p(1)=1^3-4\times1^2+3=1-4+3=1-1=0$

$n=1\\n-1=0$

So, n-1 is also factor of p(n).

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8 0

3 years ago

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The following is a linear programming formulation of a labor planning problem. There are four overlapping shifts, and management

dexar [7]

Answer:

d. 15

Step-by-step explanation:

Putting the values in the shift 2 function

X1 + X2 ≥ 15

where x1= 13, and x2=2

13+12≥ 15

15≥ 15

At least 15 workers must be assigned to the shift 2.

The LP model questions require that the constraints are satisfied.

The constraint for the shift 2 is that the number of workers must be equal or greater than 15

This can be solved using other constraint functions e.g

Putting X4= 0 in

X1 + X4 ≥ 12

gives

X1 ≥ 12

Now Putting the value X1 ≥ 12 in shift 2 constraint

X1 + X2 ≥ 15

12+ 2≥ 15

14 ≥ 15

this does not satisfy the condition so this is wrong.

Now from

X2 + X3 ≥ 16

Putting X3= 14

X2 + 14 ≥ 16

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X2 ≥ 2

Putting these in the shift 2

X1 + X2 ≥ 15

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3 years ago

how many 5-digit even numbers can be formed with the digits 1,2,3,4,5, and 6 if repetition is allowed?

Setler79 [48]

Given:

The given digits are 1,2,3,4,5, and 6.

To find:

The number of 5-digit even numbers that can be formed by using the given digits (if repetition is allowed).

Solution:

To form an even number, we need multiples of 2 at ones place.

In the given digits 2,4,6 are even number. So, the possible ways for the ones place is 3.

We have six given digits and repetition is allowed. So, the number of possible ways for each of the remaining four places is 6.

Total number of ways to form a 5 digit even number is:

$Total=6\times 6\times 6\times 6\times 3$

$Total=3888$

Therefore, total 3888 five-digit even numbers can be formed by using the given digits if repetition is allowed.

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murzikaleks [220]

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mylen [45]

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Step-by-step explanation:

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3 years ago

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