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3 years ago

12

Solve for the missing side to the nearest tenth. 19 X 51

1 answer:

telo118 [61]3 years ago

3 0

Answer:

<em>x = 30.2 units</em>

Step-by-step explanation:

<u>Trigonometric Ratios</u>

The ratios of the sides of a right triangle are called trigonometric ratios.

Selecting any of the acute angles, it has an adjacent side and an opposite side. The trigonometric ratios are defined upon those sides and the hypotenuse.

The given right triangle has an angle of measure 51° and its adjacent leg has a measure of 19 units. It's required to calculate the hypotenuse of the triangle.

We use the cosine ratio to calculate x:

$\displaystyle \cos\theta=\frac{\text{adjacent leg}}{\text{hypotenuse}}$

$\displaystyle \cos 51^\circ=\frac{19}{x}$

Solving for x:

$\displaystyle x=\frac{19}{\cos 51^\circ}$

$\displaystyle x=\frac{19}{0.6293}$

x = 30.2 units

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Answer:

If f(x) = 5x + 40, what is f(x) when x = –5 (D)

What is the inverse of the function f(x) = 2x – 10? (B)

Step-by-step explanation:

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3 years ago

Udlinius<br>6x – 12 = 6(x - 2)

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3 years ago

2x2 – 5x + 67 = 0<br> What would be your first step in completing the square for the equation above?

7nadin3 [17]

Answer:

The first step is to divide all the terms by the coefficient of $x^{2}$ which is 2.

The solutions to the quadratic equation $2x^2\:-\:5x\:+\:67\:=\:0$ are:

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

Step-by-step explanation:

Considering the equation

$2x^2\:-\:5x\:+\:67\:=\:0$

The first step is to divide all the terms by the coefficient of $x^{2}$ which is 2.

so

$\frac{2x^2-5x}{2}=\frac{-67}{2}$

$x^2-\frac{5x}{2}=-\frac{67}{2}$

Lets now solve the equation by completeing the remaining steps

Write equation in the form: $x^2+2ax+a^2=\left(x+a\right)^2$

Solving for $a$ ,

$2ax=-\frac{5}{2}x$

$a=-\frac{5}{4}$

$\mathrm{Add\:}a^2=\left(-\frac{5}{4}\right)^2\mathrm{\:to\:both\:sides}$

$x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{67}{2}+\left(-\frac{5}{4}\right)^2$

$x^2-\frac{5x}{2}+\left(-\frac{5}{4}\right)^2=-\frac{511}{16}$

Completing the square

$\left(x-\frac{5}{4}\right)^2=-\frac{511}{16}$

Since, you had required to know the first step in completing the square for the equation above, I hope you have got the point, but let me quickly solve the remaining solution.

For $f^2\left(x\right)=a$ the solution are $f\left(x\right)=\sqrt{a},\:-\sqrt{a}$

Solving

$x-\frac{5}{4}=\sqrt{-\frac{511}{16}}$

$x-\frac{5}{4}=\sqrt{-1}\sqrt{\frac{511}{16}}$

$x-\frac{5}{4}=i\sqrt{\frac{511}{16}}$ ∵ Applying imaginary number rule $\sqrt{-1}=i$

$x-\frac{5}{4}=i\frac{\sqrt{511}}{\sqrt{16}}$

$-\frac{5}{4}=i\frac{\sqrt{511}}{4}$

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4}$

Similarly, solving

$x-\frac{5}{4}=-\sqrt{-\frac{511}{16}}$

$x-\frac{5}{4}=-i\frac{\sqrt{511}}{4}$ ∵ Applying imaginary number rule $\sqrt{-1}=i$

$x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

Therefore, the solutions to the quadratic equation are:

$x=\frac{5}{4}+i\frac{\sqrt{511}}{4},\:x=\frac{5}{4}-i\frac{\sqrt{511}}{4}$

4 0

3 years ago

Do you guys get <br> thank you

Levart [38]

On the tens one shade in 3 bars representig 3/10 kn the hunderes one shade in 30 squares

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3 years ago

Read 2 more answers

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GenaCL600 [577]

Answer:

The answer is " $\neq$ "

Step-by-step explanation:

Given:

$n=97\\\\\alpha=0.10\\\\\bar{x}=63.1\\\\\sigma=4\\\\$

So, the null and alternative hypothesis is:

$H_0 : \mu =6.27\\\\H_a : \mu \neq 62.7$

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3 years ago