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Mrrafil [7]
3 years ago
5

Solve for X can you show me how to do this problem.​

Mathematics
2 answers:
sergey [27]3 years ago
7 0
Answer is 4.
Right angle= 90
50+ 10(4)= 90
saw5 [17]3 years ago
5 0

Answer:

Step-by-step explanation:

Sum of all angles of triangle = 180

<u>90 + 50</u> +10x = 180     {Combine like terms}

  <u> 140</u> + 10x = 180    {Subtract 140 from both sides}

             10x = 180 - 140

             10x = 40 {Divide both sides by 10}

                x = 40/10

x = 4

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ivann1987 [24]
<span>erry starts rowing at 2 pm from the west end of the lake, and Tao starts rowing from the east end of the lake at 2:05 pm. 
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Terry DATA:
rate = 60 meter/min ; time = x min ; distance = r*t = 60x meters
-----
Tao DATA:
rate = 40 meter/min ; time = x-5 min ; distance = r*t = 40(x-5) meters
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If they always row directly towards each other at what time will the two meet?

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Equation:
dist + dist = 2000 meters
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x = 90 minutes
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8 0
3 years ago
A)<br> Identify all x-values not in the domain of<br> y=(x² – 4x²) / (3x² - 6x – 24)
True [87]

Answer:

x ≠ 4 or -2

Step-by-step explanation:

the denominator cannot be zero, so factor the bottom equation to get the zeros and those are the domain restrictions.

3x^2 - 6x - 24 ≠ 0

3(x^2 - 2x - 8) ≠ 0          (factor out a 3)

3(x - 4)(x + 2) ≠ 0            (factor equation)

x ≠ 4, x ≠ -2                     (use zero product property to find zeros)

3 0
3 years ago
2 Points
elena55 [62]

Answer:C

Step-by-step explanation:

5 0
3 years ago
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Rudik [331]
The answer is < :) hope this helped
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3 years ago
Given the circle with the equation (x + 1)2 + y2 = 36, determine the location of each point with respect to the graph of the cir
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\bf \textit{equation of a circle}\\\\ &#10;(x- h)^2+(y- k)^2= r^2&#10;\qquad &#10;center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad &#10;radius=\stackrel{}{ r}\\\\&#10;-------------------------------\\\\&#10;(x+1)^2+y^2=36\implies [x-(\stackrel{h}{-1})]^2+[y-\stackrel{k}{0}]^2=\stackrel{r}{6^2}~~~~&#10;\begin{cases}&#10;\stackrel{center}{(-1,0)}\\&#10;\stackrel{radius}{6}&#10;\end{cases}

so, that's the equation of the circle, and that's its center, any point "ON" the circle, namely on its circumference, will have a distance to the center of 6 units, since that's the radius.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}&#10;\\\\&#10;(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad &#10;A(\stackrel{x_2}{-1}~,~\stackrel{y_2}{1})\qquad \qquad &#10;%  distance value&#10;d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}&#10;\\\\\\&#10;\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(1-0)^2}\implies d=\sqrt{(-1+1)^2+1^2}&#10;\\\\\\&#10;d=\sqrt{0+1}\implies d=1

well, the distance from the center to A is 1, namely is "inside the circle".

\bf ~~~~~~~~~~~~\textit{distance between 2 points}&#10;\\\\&#10;(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad &#10;B(\stackrel{x_2}{-1}~,~\stackrel{y_2}{6})\\\\\\&#10;\stackrel{distance}{d}=\sqrt{[-1-(-1)]^2+(6-0)^2}\implies d=\sqrt{(-1+1)^2+6^2}&#10;\\\\\\&#10;d=\sqrt{0+36}\implies d=6

notice, the distance to B is exactly 6, and you know what that means.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}&#10;\\\\&#10;(\stackrel{x_1}{-1}~,~\stackrel{y_1}{0})\qquad &#10;C(\stackrel{x_2}{4}~,~\stackrel{y_2}{-8})&#10;\\\\\\&#10;\stackrel{distance}{d}=\sqrt{[4-(-1)]^2+[-8-0]^2}\implies d=\sqrt{(4+1)^2+(-8)^2}&#10;\\\\\\&#10;d=\sqrt{25+64}\implies d=\sqrt{89}\implies d\approx 9.43398

notice, C is farther than the radius 6, meaning is outside the circle, hiking about on the plane.
3 0
3 years ago
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