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3 years ago

Solve the system by the elimination method. 2a + 3b = 6 5a + 2b - 4 = 0

2 answers:

5 0

2a + 3b = 6 --> Equation 1
5a + 2b = 4 --> Equation 2
Equation 1 * 5 --> 10a + 15b = 30
Equation 2 * 2 --> 10a + 4b = 8
Equation 1 - Equation 2 --> 11b = 22 --> b = 2
Sub b=2 in Equation 1 (or 2, either works):
2a + 3b = 6
2a + 3(2) = 6
2a + 6 = 6
2a = 0
a = 0

b = 2, a = 0

dem82 [27]3 years ago

5 0

Make in form
ax+by=c

add 4 to both sides in 2nd equaiton

5a+2b=4

now
elimiate a variable
I choose b
multiply top equation by -2 and 2nd by 3
add

-4a-6b=-12
15a+6b=12 +
11a+0b=0

11a=0
a=0
nice

sub back
5a+2b=4
5(0)+2b=4
2b=4
divide 2
b=2

a=0
b=2
if in form (a,b)
(0,2)

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3 years ago

2 tan 30° II 1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago

I really don’t get it

posledela

you just add 8 + 8 + 8 it equals 24. rounded it will be 20

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3 years ago

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Answer:

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