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Julli [10]
4 years ago
10

There are 20 dowels. All bevels are dowels. All dowels are swivels. There are 12 bevels. There are 30 swivels. How many swivels

are not bevels or dowels?

Mathematics
2 answers:
Zina [86]4 years ago
7 0

Answer:

10 swivels are neither bevels or dowels

Step-by-step explanation:

See the attached for explanation

spin [16.1K]4 years ago
4 0

Answer:

10

Step-by-step explanation:

Given that all bevels are dowels, where there are  12 bevels and 20 dowels, it means there are 12 bevels that are dowels and 8 dowels that are not bevels.

Furthermore, if all dowels are swivels, and there are 30 swivels then the number of swivels that are neither bevels nor dowels

= 30 - 20

= 10

There are 10 swivels that are not bevels or dowels.

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Write each of the below number in an ascending order.<br> 1.5 1.75 2.07 2.8
34kurt

Answer:

1.5, 1.75, 2.07, 2.8

Step-by-step explanation:

Ascending order means to have the smallest value first and going up to the largest.

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2 years ago
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nydimaria [60]

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2 years ago
Verify sine law by taking triangle in 4 quadrant<br>Explain with figure.<br>​
Ksivusya [100]

Proof of the Law of Sines

The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)

a

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For more see Law of Sines.

Acute triangles

Draw the altitude h from the vertex A of the triangle

From the definition of the sine function

 sin  B =

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Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

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Repeat the above, this time with the altitude drawn from point B

Using a similar method it can be shown that in this case

c

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Combining (4) and (5) :

a

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- Q.E.D

Obtuse Triangles

The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.

First the interior altitude. This is the same as the proof for acute triangles above.

Draw the altitude h from the vertex A of the triangle

 sin  B =

h

c

      a n d          sin  C =

h

b

or

h = c  sin  B       a n d         h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Draw the second altitude h from B. This requires extending the side b:

The angles BAC and BAK are supplementary, so the sine of both are the same.

(see Supplementary angles trig identities)

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 sin  A =

h

c

or

h = c  sin  A

In the larger triangle CBK

 sin  C =

h

a

or

h = a  sin  C

From (6) and (7) since they are both equal to h

c  sin  A = a  sin  C

Dividing through by sinA then sinC:

a

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Combining (4) and (9):

a

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7 0
3 years ago
Please help step by step!!!
Lesechka [4]

Answer:

-7

Step-by-step explanation:

We first are going to flip the equation around.

So instead of 9 times x = -36 we'll do 36/9. 36/9 = -7.

Also you will kind of know it will be a negative number from the beginning since it's -36.

I don't know if that last part makes any sense, but I hope this helps you!

5 0
3 years ago
Read 2 more answers
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Harman [31]

Answer:

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Since the differences are not the same, Mrs Bailey must first perform a (slope) multiplication by a factor of 30/40 or 3/4

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8 0
3 years ago
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