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andrey2020 [161]
3 years ago
11

Solve set of each equation 2|3x-2|= 14

Mathematics
1 answer:
avanturin [10]3 years ago
6 0
<h3>Solution:</h3>

2(3x - 2) = 14 \\  =  > 6x - 4 = 14 \\  =  > 6x = 14 + 4 \\  =  > 6x = 18 \\  =  > x =  \frac{18}{6}  \\  =  > x = 3

<h3>Answer:</h3>

The answer is 3.

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3.33 is the right answer
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What is the value of tan(60°)? One-half StartRoot 3 EndRoot StartFraction StartRoot 3 EndRoot Over 2 EndFraction StartFraction 1
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Answer:

\sqrt3

Step-by-step explanation:

To find:

The value of tan60^\circ = ?

Solution:

Kindly consider the equilateral \triangle ABC as attached in the answer area.

Let the side of triangle = a units

Let us draw the perpendicular from vertex A to side BC.

It will divide the side BC in two equal parts.

i.e. BD = DC = \frac{a}{2}

Using Pythagorean Theorem in \triangle ABD:

\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}

Side AD = \frac{\sqrt3}{2}a

Using Trigonometric ratio:

tan\theta = \dfrac{Perpendicular}{Base}

tanB = \dfrac{AD}{BD}

Putting the values of AD and BD:

tan60^\circ=\dfrac{\frac{\sqrt3}{2}a}{\frac{1}{2}a}\\\Rightarrow tan60^\circ = \bold{\sqrt3}

5 0
3 years ago
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Nick and Adam work out together neck weighs 160 pounds is and is gaining 30 pounds per week Adam weighs 195 pounds and is losing
muminat

Answer:

(N+n)=(A+a)

Step-by-step explanation:

Let N represent Nick's current weight

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Answer:

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therefore:

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2:

- 4 {a}^{5} (6 {a}^{5} ) \\  = ( - 4 \times 6)( {a}^{5}  \times  {a}^{5} ) \\  =  - 24  \{{a}^{(5 + 5)}  \} \\  = { \boxed{ \boxed{ - 24 {a}^{10} }}}

3:

( - 7 {a}^{4} b {c}^{3} )(5a {b}^{4}  {c}^{2} ) \\  = ( - 7 \times 5)( {a}^{4}  \times a)(b \times  {b}^{4})  \times ( {c}^{3}  \times  {c}^{2} ) \\  =  - 35 {a}^{5}  {b}^{5 }  {c}^{5}  \\  = { \boxed{ \boxed{ - 35 {(abc)}^{5} }}}

4:

\frac{ {8}^{15} }{ {8}^{13} }

from law of indices:

\frac{a {}^{n} }{ {a}^{m} }  =  {a}^{(n - m) }  \\

therefore:

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follow the procedure for application to other questions.

<em>Any</em><em> </em><em>question</em><em>,</em><em> </em><em>be</em><em> </em><em>free</em><em> </em><em>to</em><em> </em><em>ask</em><em>.</em><em> </em>Thank you …

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