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pav-90 [236]
2 years ago
10

(Pls help) I need help finding x!

Mathematics
1 answer:
Ray Of Light [21]2 years ago
8 0

Answer:

x ≈ 46°

Step-by-step explanation:

Since the triangle is right use the sine ratio to solve for x

sinx = \frac{opposite}{hypotenuse} = \frac{10}{14}, hence

x = sin^{-1} ( \frac{10}{14} ) ≈ 46°

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Simplify the expression 6[8+8](3+7)] ?
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Answer:

960

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6[8+8](3+7)

Parentheses first

6[16] (10)

Then multiply from left to right

96*10

960

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3 years ago
Geometry Homework Question ​
mash [69]

Answer:

Distance is 8.485

Midpoint is (3,4)

Step-by-step explanation:

Use the points (-1,7) and (5,1)

Insert them into the equation and solve for D

D=\sqrt{x} ((5-(-1))^2 + (1-7)^2)\\D=\sqrt{x} ((6)^2 + (-6)^2)\\D=\sqrt{x} (36+36)\\D=\sqrt{x} (72)\\D=8.485

Insert points into second equation to find midpoint

(\frac{_1+5}{2} , \frac{7+1}{2}) \\(\frac{6}{2} ,\frac{8}{2} )\\(3,4)

7 0
3 years ago
Find the point P on the line y=2x that is closest to the point (20,0) .what is the least distance between pans (20,0)?
Vsevolod [243]
The closest point on a line, to another point, will be a point that's on a normal of that line, or a line that is perpendicular to it, notice picture below

so, y = 2x, has a slope of 2, a perpendicular line to it, will have a slope of negative reciprocal that, or \bf 2\qquad negative\implies -2\qquad reciprocal\implies \cfrac{1}{-2}\implies -\cfrac{1}{2}

so, we know that line passes through the point 20,0, and has a slope of -1/2

if we plug that in the point-slope form, we get \bf y-0=-\cfrac{1}{2}(x-20)\implies y=-\cfrac{1}{2}x+10

now, the point that's on 2x and is also on that perpendicular line, is the closest to 20,0 from 2x, thus, is where both graphs intersect, as you can see in the graph

thus  \bf 2x=-\cfrac{1}{2}x+10  solve for "x'

------------------------------------------------------------

not sure on the 2nd part, but sounds like, what's the distance from that point to 20,0, well, if that's the case, just use the distance equation

\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ \square }}\quad ,&{{ \square }})\quad 
%  (c,d)
&({{ \square }}\quad ,&{{ \square }})
\end{array}\qquad 
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}


6 0
2 years ago
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