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2 years ago

I WILL GIVE BRAINLIEST 50 POINTS Find x. Round your answer to the nearest tenth of a degree.

Mathematics

2 answers:

sveta [45]2 years ago

8 0

Answer:

c = 30.8 (nearest tenth)

Step-by-step explanation:

Pythagoras' Theorem: a² + b² = c²

(where a and b are the legs, and c is the hypotenuse, of a right triangle)

Given:

a = 21
b = 22.5

Substitute given values into the formula and solve for c:

⇒ 21² + 22.5² = c²

⇒ 947.25 = c²

⇒ c = ±√(947.25)

⇒ c = ±30.8 (nearest tenth)

As distance is positive, c = 30.8 (nearest tenth) only

Angelina_Jolie [31]2 years ago

3 0

$\\ \rm\rightarrowtail tan47=\dfrac{b}{21}$

$\\ \rm\rightarrowtail b=21tan47$

$\\ \rm\rightarrowtail b=22.5$

$\\ \rm\rightarrowtail cos47=\dfrac{21}{c}$

$\\ \rm\rightarrowtail c=21/cos47$

$\\ \rm\rightarrowtail c=30.88$

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Step-by-step explanation:

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2 years ago

Simplify radical expression

Sedbober [7]

Answer:

the right answer is option D. 0.5b

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3 years ago

Read 2 more answers

Explain how to get that answer!!

ra1l [238]

We need to simplify $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$

First lets factor $\sqrt{14x^3}$

$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$
$\sqrt{14} = \sqrt{2} \sqrt{7}$ by applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$
$\sqrt{x^3} = x^{3/2}$ By applying the radical rule $\sqrt[n]{x^m} = x^{m/n}$

So
$\sqrt{14x^3}$ = $\sqrt{14} \sqrt{x^3}$ = $\sqrt{2} \sqrt{7}x^{3/2}$

Now let's factor $\sqrt{18x}$
By applying the radical rule $\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b}$ ,
$\sqrt{18x} = \sqrt{18} \sqrt{x}$
$\sqrt{18} = \sqrt{2} * 3$

So $\sqrt{18x}$ = $\sqrt{2}*3 \sqrt{x}$

So $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x} }$

We know that $\sqrt[n]{x} = x^{1/n}$ so $\sqrt{x} = x^{1/2}$

We now have $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3 \sqrt{x}} = \frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}}$

We know that $\frac{x^a}{x^b} = x^{a-b}$
So $\frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2} = x$

We now got $\frac{ \sqrt{2} \sqrt{7} x^{3/2} }{ \sqrt{2}*3x^{1/2}} = \frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3}
$

We can notice that the numerator and the denominator both got √2 in a multiplication, so we can simplify them, and we get:
$\frac{ \sqrt{2} \sqrt{7} x }{ \sqrt{2}*3} = \frac{ \sqrt{7}x }{3}$

All in All, we get $\frac{ \sqrt{14x^3} }{ \sqrt{18x} }$ = $\frac{ \sqrt{7}x }{3}$

Hope this helps! :D

6 0

3 years ago

The scatter plot shows which type of correlation?

andre [41]

Answer:

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Step-by-step explanation:

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8 0

3 years ago

How do I do this? Please provide steps.

matrenka [14]

Answer:

3/5

Step-by-step explanation:

Soh Cah Toa

In some trigonometry classes, this acronym is very important to solving questions like this.

Why?

It tells us the right triangle-definition of these trigonometry functions.

We have that cosine of an angle is equal to tge side that is adjacent to it over the hypotenuse.

So here we are asked to find cos(B).

Lets look at triangle respect to the angle B

The measurement of the side that is opposite is 64.

The measurement of the side that is adjacent is 48.

The measurement of the hypotenuse is 80.

So cos(B)=48/80.

Let's reduce. 48 and 80 have a common factor of 8 so divide numerator and dexter by 8. This gives us:

cos(B)=6/10.

One more step in reducing. Both factors are even so cos(B)=3/5.

6 0

3 years ago