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3 years ago

9

Solve: log(2)(3x + 8) = 5 Which is an equivalent equation?

2 answers:

Rzqust [24]3 years ago

8 0

You can just simplify to make an equivalent equation:

(2)(3x + 8) = 5 ⇒ 6x + 16 = 5

Answer:

$\boxed{\bf~6x + 16 = 5}$

Hope it helped,

BioTeacher101

<em>(If you have any questions feel free to ask them in the comments)</em>

Debora [2.8K]3 years ago

5 0

<u>Answer</u>

x = 8

<u>Explanation</u>

㏒₂ (3x + 8) = 5

5 =㏒₂ 2⁵.

∴ ㏒₂ (3x + 8) = ㏒₂ 2⁵

If the right side of the equation above is equal to the left side, then it is true to say that;

(3x + 8) = 2⁵

3x + 8 = 32

3x = 32 - 8

3x = 24

x = 24/3

= 8

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A running circuit is in the shape of a triangle with lengths of 6km, 6.5km and 7km. What are the sizes of the angles (in minutes

Rudik [331]

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to $180^{o}$ . The <em>measures</em> of the internal <u>angles</u> of the <u>triangle</u> given in the question are A = $52.6^{o}$ , B = $59.4^{o}$ , and C = $68^{o}$ .

A <u>triangle</u> is an example of a class of <em>figures</em> referred to as <em>plane shapes</em>. It has <u>three</u> straight <u>sides</u> and <u>three</u> internal <u>angles</u> which sum up to $180^{o}$ .

Considering the given question, let the <u>sides</u> of the triangle be: a = 6 km, b = 6.5 km, and c = 7 km.

Apply the <em>Cosine rule</em> to have:

$c^{2}$ = $a^{2}$ + $b^{2}$ - 2ab Cos C

So that;

$7^{2}$ = $6^{2}$ + $(6.5)^{2}$ - 2(6 * 6.5) Cos C

49 = 36 + 42.25 - 78Cos C

78 Cos C = 78.25 - 49

= 29.25

Cos C = $\frac{29.25}{78}$

= 0.375

C = $Cos^{-1}$ 0.375

= 67.9757

C = $68^{o}$

Apply the <em>Sine rule</em> to determine the <u>value</u> of B,

$\frac{b}{Sin B}$ = $\frac{c}{Sin C}$

$\frac{6.5}{Sin B}$ = $\frac{7}{Sin 68}$

SIn B = $\frac{6.5 *Sin 68}{7}$

= 0.861

B = $Sin^{-1}$ 0.861

= 59.43

B = $59.4^{o}$

Thus to determine the value of A, we have;

A + B + C = $180^{o}$

A + $59.4^{o}$ + $68^{o}$ = $180^{o}$

A = $180^{o}$ - 127.4

= 52.6

A = $52.6^{o}$

Therefore the <u>sizes</u> of the <em>internal angles</em> of the triangle are: A = $52.6^{o}$ , B = $59.4^{o}$ , and C = $68^{o}$ .

For more clarifications on applications of the Sine and Cosine rules, visit: brainly.com/question/14660814

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