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MrMuchimi
3 years ago
13

What are the solutions to the quadratic equation 4(x + 2)2 = 36

Mathematics
2 answers:
Alina [70]3 years ago
8 0

Answer:

C. x = -5, and x = 1

Step-by-step explanation:

You have the equation 4(x + 2)2 = 36

You then isolate 4 out of the equation using division on both sides

You are left with (x+2)2 = 9

Next, take square roots

\sqrt({x} +2) ^{2} = \sqrt{9}

This leaves you with x + 2 ± 9

So, x = -2 + 3 = 1

or

x = -2 - 3 = -5

kogti [31]3 years ago
3 0

Answer:

x = -5 and x = 1

Step-by-step explanation:

4( x2 + 4x + 4 ) = 36

4x2 + 16x + 16 = 36

4x2 + 16x - 20 = 0

x2 + 4x - 5 = 0

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Bas_tet [7]
Hey there! I’m happy to help!

We see that the two equal sides are longer than the base. This means that our a will be larger than b. The problem says that one of the longer sides is 6.3. Since the two longer sides are equal, that means that both longer sides are 6.3 cm. Therefore, we can simply subtract these two from the perimeter to find the shorter side, which is the base.

15.7-6.3-6.3=3.1

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I hope that this helps!
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3 years ago
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Step-by-step explanation:

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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

 sin  C

7 0
2 years ago
What statement is true? 1) 5 is a whole and not a integer. 2) 5 is a integer and not a whole. 3) 5 isn't a integer or whole 4) 5
Lorico [155]

Answer:

4

Step-by-step explanation:

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alexandr402 [8]

Answer:

(2-c,y)

Step-by-step explanation:

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3 years ago
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