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2 years ago

Help help help help plssssss

Mathematics

1 answer:

dolphi86 [110]2 years ago

4 0

Answer:

${ \tt{x + (x + 6) + 114 = 360}} \\ { \tt{ \blue{sum \: of \: exterior \: angles \: is \: 360 \degree}}} \\ { \tt{2x + 120 = 360}} \\ { \tt{2x = 240}} \\ { \tt{x = 120}}$

Angles are 120°, 126°, and 114°

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A book has a front and back and 400 pages .

a_sh-v [17]

The minimum thickness of the book would simply be the sum of the thickness of the two cover pages and all the pages in between. That is:

minimum thickness = 0.9 mm * 2 + 0.13 mm * 400

minimum thickness = 53.8 mm

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3 years ago

What is the average rate of change of f(x) over the interval [6, 8]?

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3 years ago

Which number is additive inverse of -4 1/4

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3 years ago

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3 years ago

Find the equation for the circle with a diameter whose endpoints are (3,-1) and (-2,-1)

cricket20 [7]

Answer:

The equation for the circle is: $(x-\frac{1}{2} )^2 + y^2 = (\frac{5}{2} )^2$

Step-by-step explanation:

Here, let us assume the two end points of the diameter are:

P (3,-1) and Q (-2,-1)

Now, as we knew diameter is the chord that passer through the center of the circle.

⇒ Mid point of DIAMETER = Center coordinates of the Circle

Let us assume O(x,y) is the center of the line segment PQ.

So, by MID POINT FORMULA:

$(x,y)= (\frac{3 -2}{2} , \frac{-1-1}{2} ) =( \frac{1}{2},\frac{-2}{2}) \\\implies (x,y) =( 0.5 , -1)$

⇒ The center coordinates of the circle = O(0.5,-1) ...... (1)

Now, RADIUS = Half of DIAMETER

Using DISTANCE FORMULA:

$PQ = \sqrt{(3-(-2))^2 + (-1 - (-1))^2} = \sqrt{(5)^2 + 0} = 5$

So, Radius = 5/2 = 2.5 units

Now, the equation of circle with radius r and center coordinate ( h,k) is given as: $(x-h)^2 + (y-k)^2 = r^2$

Substitute r = 2.5 and (h,k) = (0.5,0) we get:

$(x-0.5)^2 + (y-0)^2 = (2.5)^2\\\implies (x-\frac{1}{2} )^2 + y^2 = (\frac{5}{2} )^2$

Hence the equation for the circle is: $(x-\frac{1}{2} )^2 + y^2 = (\frac{5}{2} )^2$

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4 years ago