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Serjik [45]
3 years ago
11

3 over 27 equals n over 72, I have to figure out n

Mathematics
2 answers:
Olin [163]3 years ago
6 0
3/27 = n/72

\frac{3}{27}      \frac{n}{72}

mezya [45]3 years ago
3 0
Here's an equation for you
3 \times 72 = 27n
Can you solve it from here?
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Consider the function f(x) = x^{2/3} on the interval [-1,8].
zhannawk [14.2K]
The right option is (c). The Mean Value Theorem does not apply since f(x) is not differentiable on (-1,8). Because the derivative of f is 2/3 x^(-1/3)  which is undefined at x=0.So, that would imply that it is not differentiable along the interval.
5 0
3 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

 sin  A

=

b

 sin  B

=

c

 sin  C

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 =

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

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

 sin  C

=

a

 sin  A

Combining (4) and (5) :

a

 sin  A

=

b

 sin  B

=

c

 sin  C

- 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)

Angle A is BAC, so

 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

 sin  A

=

c

 sin  C

Combining (4) and (9):

a

 sin  A

=

b

 sin  B

=

c

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7 0
3 years ago
State if the two triangles are congruent. If they are, state how you know.
Gwar [14]

Answer:

The triangles can be congruent.

Step-by-step explanation:

They are congruent if proven by SSS: 2 sides are clearly stated that they are congruent due to the marks they have.

The last side can be congruent if the diagonals are congruent in length by proving.

They can also be congeuent due to SAS because there is gonna be alternate interior angles due to the transversal.

4 0
3 years ago
Use 4 terms of the series to approximate :
liraira [26]

from \: trapezium \: rule \\ n = 4 \\ h =  \frac{(1 - ( - 5))}{4}  =  \frac{3}{2}  \\ from \: the \: pic \:  \\ approximate \: value \: is \: 5.409

8 0
3 years ago
The perimeter of the polygon above is
Rama09 [41]

Answer:

80 cm

Step-by-step explanation:

5 points in a star, each point has a perimeter of 16,

16 x 5 = 80

or just legit add all the 8's

DONT FORGET TO INCLUDE THE UNITS

4 0
3 years ago
Read 2 more answers
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