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3 years ago

a segment has and endpoint at (-6, 8). The midpoint is at (-6, 2). What are the coordinates of the other endpoint?

Mathematics

1 answer:

Montano1993 [528]3 years ago

8 0

Answer:

(-6,-4)

Step-by-step explanation:

The first endpoint of the line is (-6,8), we can call

x_1 = -6

and

y_1 = 8

Let the last endpoint have coordinates (x_2,y_2)

Also, the midpoint formula is:

(x_1+x_2)/2 , (y_1+y_2)/2

Now, plugging these values is the formula, we get:

(-6+x_2)/2 = -6

-6+x_2=-12

x_2=-12+6 = -6

x_2 = -6

Also

(8+y_2)/2=2

8+y_2=4

y_2=4-8=-4

y_2 = -4

The coordinates of the other endpoint is (-6,-4)

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Cos (-Ø)= √3/3 sinØ <0 what is the value of sinØ?

Inga [223]

Answer:

sin∅= - √6/3

Step-by-step explanation:

Trigonometric function cos is even function which mean:

cos (-∅) = cos ∅

Trigonometric function sin is odd function which mean:

sin (-∅) = - sin∅ => sin∅ < 0

The basic trigonometric equation reads in this case with angle marked ∅:

sin²∅ + cos²∅ = 1 => sin²∅ = 1 - cos²∅ => sin²∅ = 1 - ( √3/3 )² =>

sin²∅ = 1 - (3/9) => sin²∅ = 1 - (1/3) = (3/3) - (1/3) = 2/3 => sin∅ = - √(2/3) =>

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3 years ago

8 3/4 +n+ 2 3/8 +23 1/8 <br> solve for n

bezimeni [28]

8 3/4+n+2 3/8+ 23 1/8
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70/8 +n+ 204/8
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4 0

3 years ago

Rewrite the expression in the form... (PLZ HELP QUICK I'LL GIVE BRAINLIEST)

ivanzaharov [21]

Answer: $3*z^\frac{2}{9}$

Step-by-step explanation:

The exponent 1/3 is the same thing as cube rooting the expression.

$\sqrt[3]{27\sqrt[3]{z^2} }$

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3 years ago

A rocket is launched from the top of a 99-foot cliff with an initial velocity of 122 ft/s.

Harman [31]

Remember that c is the initial height. Since we the rocket is in a 99-foot cliff, c=99. Also, we know that the velocity of the rocket is 122 ft/s; therefore v=122
Lets replace the values into the the vertical motion formula to get:
$0=-16 t^{2} +122t+99$
Notice that the rocket hits the ground at the bottom of the cliff, which means that the final height is 99-foot bellow its original position; therefore, our final height will be h=-99
Lets replace this into our equation to get:
$-99=-16 t^{2} +122t+99$
$-16 t^{2} +122+198=0$

Now we can apply the quadratic formula $t= \frac{-b+or- \sqrt{ b^{2} -4ac} }{2a}$ where a=-16, b=122, and c=198
$t= \frac{-122+or- \sqrt{ 122^{2}-(4)(-16)(198) } }{(2)(-16)}$
$t= \frac{-122+ \sqrt{27556} }{-32}$ or $t= \frac{-122- \sqrt{27556} }{-32}$
$t= \frac{-122+166}{-32}$ or $t= \frac{-122-166}{-32}$
$t= \frac{-11}{8}$ or $t=9$

Since the time can't be negative, we can conclude that the rocket hits the ground after 9 seconds.

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3 years ago

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Makovka662 [10]

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