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2 years ago

Find the value of x

Mathematics

2 answers:

Sunny_sXe [5.5K]2 years ago

7 0

Answer:

19.1

Step-by-step explanation:

Let's assign some variables to this triangle.

angle A = 63°

angle B = unknown

angle C = 71°

side a = 18

side b = y

side c = x

By the law of sins, applicable to ALL triangles, we have:

$\frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c}$

Let's then take the equation for angles A and C, in order to isolate the c (which is our x to find).

we then have

$c = \frac{sin(C) * a}{sin(A)} = \frac{sin(71) * 18}{sin(63)} = 19.10$

So, the answer is the second one.

fenix001 [56]2 years ago

5 0

Answer:

The correct answer is second option 19.1

Step-by-step explanation:

From the figure we can see that, a triangle.

Le z be the perpendicular distance

<u>To find the value of x</u>

Using trigonometric ratio we can write

Sin 71 = Z/18

Z = 18 * Sin 71 = 17.01

Sin 63 = Z/x

x = Z/Sin 63 = 17.01/0.891 = 19.1

Therefore the correct answer is optin d. 19.1

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What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?

astraxan [27]

ANSWER

The value of the expression is
$- 1$

EXPLANATION

Method 1: Rewrite as product of
${i}^{2}$

The expression given to us is,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4}$

We use the fact that
${i}^{2} = - 1$
to simplify the above expression.

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{1} \times {i}^{3} \times {i}^{2} \times {i}^{4}$

This implies,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2}$

We substitute to obtain,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times - 1 \times - 1 \times - 1\times - 1 \times - 1$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times 1 \times 1 \times - 1 = - 1$

Method 2: Use indices to solve.

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0 + 1 + 2 + 3 + 4}$

This implies that,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{10}$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( {{i}^{2}} )^{5}$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( - 1 )^{5} = - 1$

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3 years ago

Read 2 more answers

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