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3 years ago

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Use the figure to find the trigonometric ratio below. Express the answer as a decimal rounded to the nearest ten-thousandth.

2 answers:

Ira Lisetskai [31]3 years ago

8 0

Answer:

The correct option is 1.

Step-by-step explanation:

Given information: AD = 25, CD = 5, DB = 1 and CD⊥AB.

According to the Pythagoras theorem,

$hypotenuse^2=base^2+perpendicular^2$

In triangle BCD,

$CB^2=DB^2+CD^2$

$CB^2=1^2+5^2$

$CB^2=26$

Taking square root both sides.

$CB=\sqrt{26}$

In a right angled triangle,

$\sin\theta=\frac{opposite}{hypotenuse}$

$\sin B=\frac{CD}{CB}$

$\sin B=\frac{5}{\sqrt{26}}$

$\sin B=0.980580675691$

$\sin B\approx 0.9806$

Therefore the correct option is 1.

OlgaM077 [116]3 years ago

6 0

Answer:

0.9806 is the correct answer.

Step-by-step explanation:

You might be interested in

For what values of the variables are the following expressions defined? 1. 5y+2 2. 18/y 3. 1/x+7 4. 2b/10−b Example: X>7

Serga [27]

Answer:

1. All real numbers

2. All real numbers except y = 0

3. All real numbers except x = -7

4. All real numbers except b = 10

Step-by-step explanation:

For any function to be defined at a particular value, it should not be <em>approaching to a value </em> $\infty$ <em> or it should not give us the </em> $\frac{0}{0}$ <em> (zero by zero) form </em> when the input is given to the function.

The value of function will depend on the denominator.

Now, let us consider the given functions one by one:

1. 5y+2

Here denominator is 1. So, it can not attain a value $\infty$ or $\frac{0}{0}$ <em> (zero by zero) form </em>

So, for all real numbers, the function is defined.

$2.\ \dfrac{18}{y}$

At y = 0, the value

$At\ y =0, \dfrac{18}{y} \rightarrow \infty$

So, the given function is <em>defined for all real numbers except y = 0</em>

<em></em>

<em></em> $3.\ \dfrac{1}{x+7}$ <em></em>

Let us consider denominator:

x + 7 can be zero at a value x = -7

$At\ x =-7, \dfrac{1}{x+7} \rightarrow \infty$

So, the given function is <em>defined for all real numbers except x = -7</em>

<em></em>

$4.\ \dfrac{2b}{10-b}$

Let us consider denominator:

10-b can be zero at a value b = 10

$At\ b =10, \dfrac{2b}{10-b} \rightarrow \infty$

So, the given function is <em>defined for all real numbers except b = 10</em>

<em></em>

7 0

3 years ago

Apply the distributive property to write an equivalent expression.

aleksley [76]

Answer:

6x + 48

Step-by-step explanation:

multiply each term inside the () by 6

6x + 6*8

7 0

3 years ago

Solve the proportion for x

S_A_V [24]

Answer:

8

Step-by-step explanation:

45 divided by 5 is 9 and what you do on the to you do on the bottom which means you would do 40 divided by 5 which you would get 8 as your answer.

7 0

3 years ago

You are 5 feet tall and cast an 8-foot shadow. A lamppost nearby casts a shadow that is 20 feet. Which equation can you use to s

aleksandrvk [35]

I believe the answer is the first option. Tell me if i'm wrong. :)

3 0

3 years ago

Read 2 more answers

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

<em>C.</em> $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em> $(5-\frac{1}{2})^6$ <em />

3 0

3 years ago