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nalin [4]
2 years ago
12

PLEASE HELP!! use trigonometry to solve for the missing side of x of the right triangle

Mathematics
2 answers:
Rina8888 [55]2 years ago
5 0

Answer:

x = 10.7775

Step-by-step explanation:

We can see that we need to use sin∅ to help solve for <em>x</em>:

sin56° = x/13

13sin56° = x

x = 10.7775

Grace [21]2 years ago
5 0

Answer:

10.77

Step-by-step explanation:

Pretty easy stuff, basically you use the sin ratio to solve this.

so it would be sin(56) = x/13

then you multiply 13 to sin(56)

and get 10.77.

Hope this helped :)

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Find the type and number of solution for g(x) = x2 -14x = -50
inysia [295]

Answer:

Step-by-step explanation:

x² -14x+50= 0

this equation has 2 solutions because is a quadratic

the solurions are imaginary roots because the discriminant is less then 0

b²-4ac = (-14)²-4*1*50 = 194-200= -6

to find the actual roots use the quadratic formula

3 0
2 years ago
A flagpole cast a shadow 16 feet long at the same time a pole 9 feet high casts a shadow 6feet long what is the height in feet o
Doss [256]
The pole would be 24 feet high.
X/16=9/6
X=24
7 0
3 years ago
Question 18
kobusy [5.1K]

Answer:

Undefined

Step-by-step explanation:

y=-7 has a slope of 0, since it is a horizontal line. The only types of lines that are perpendicular to horizontal lines are vertical lines, which are usually in the form x=a (where a is any real number). Vertical lines have an undefined slope, so the answer is undefined. Hope this helps!

6 0
2 years ago
6y-6y-12y=please help me
Ray Of Light [21]
Hello.

The answer is 

-12y

Combine Like Terms:<span>=<span><span><span>6y</span>+<span>−<span>6y</span></span></span>+<span>−<span>12y</span></span></span></span><span>=<span>(<span><span><span>6y</span>+<span>−<span>6y</span></span></span>+<span>−<span>12y</span></span></span>)</span></span><span>=<span>−<span>12<span>y

Have a nice day</span></span></span></span>
7 0
3 years ago
Read 2 more answers
The vertices of triangle RST are R (-7, 5), S(17, 5), T(5, 0). What is the perimeter of triangle RST?
MaRussiya [10]

~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ R(\stackrel{x_1}{-7}~,~\stackrel{y_1}{5})\qquad S(\stackrel{x_2}{17}~,~\stackrel{y_2}{5}) ~\hfill RS=\sqrt{(~~ 17- (-7)~~)^2 + (~~ 5- 5~~)^2} \\\\\\ ~\hfill RS=\sqrt{( 24)^2 + ( 0)^2}\implies \boxed{RS=24} \\\\\\ S(\stackrel{x_1}{17}~,~\stackrel{y_1}{5})\qquad T(\stackrel{x_2}{5}~,~\stackrel{y_2}{0}) ~\hfill ST=\sqrt{(~~ 5- 17~~)^2 + (~~ 0- 5~~)^2}

~\hfill ST=\sqrt{( -12)^2 + ( -5)^2}\implies \boxed{ST=13} \\\\\\ T(\stackrel{x_1}{5}~,~\stackrel{y_1}{0})\qquad R(\stackrel{x_2}{-7}~,~\stackrel{y_2}{5}) ~\hfill TR=\sqrt{(~~ -7- 5~~)^2 + (~~ 5- 0~~)^2} \\\\\\ ~\hfill TR=\sqrt{( -12)^2 + (5)^2}\implies \boxed{TR=13} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{\textit{\LARGE perimeter}}{24~~ + ~~13~~ + ~~13\implies \text{\LARGE 50}}~\hfill

3 0
1 year ago
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