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klasskru [66]
1 year ago
11

If a line that passes through the

Mathematics
1 answer:
4vir4ik [10]1 year ago
4 0
Gradient of line= (y2-y1)/(x2-x1)
Gradient of line= (10-(-6))/(8-4)
Gradient of line= 16/4
Gradient of line= 4

Y-intercept
Y=4x+c
When x=8, y=10
10=32+c
c= -22

Equation of line
Y=4x-22

When x=b, y=2b
Y=4x-22
2b=4b-22
22=2b
b=11
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Find (if possible) the complement and the supplement of each angle. (if not possible, enter impossible.) (a) π 10 complement rad
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In this question, we have to find the complement and supplement of the given angles .

Complementary angles are those angles whose sum is 90 degree and supplementary angles are those angles whose sum is 180 degree.

So to find the complement and supplement angles, we need to subtract the given angles from pi/2 and pi respectively .

a.

\frac{ \pi}{10}&#10;\\&#10;complement  \ angle = \frac{ \pi}{2} - \frac{ \pi}{10}     \\                             = \frac{4 \pi}{10} = \frac{ 2 \pi}{5}&#10;\\&#10;Supplement  \ angle = \pi - \frac{ \pi}{10} = \frac{9 \pi}{10}

b.

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What is the least amount of wrapping paper needed to wrap a gift box that measures 8 inches by 8 inches by 10 inches. Explain
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448 square inches.

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Step-by-step explanation:

distance AB is (using Pythagoras with the coordinate differences between both points as sides and the distance as Hypotenuse = baseline of the triangle, the side opposite of the 90 degree angle):

distance² = (5 - -10)² + (7-2)² = 15² + 5² = 225 + 25 = 250

distance = sqrt(250)

to split this distance into a 3/2 ratio, we need actually 3+2=5 equal parts. and then AM gets 3 off these parts, and MB gets 2 of these parts.

so,

distance/5 = sqrt(250)/5 = sqrt(250/25) = sqrt(10)

so,

AM = 3×sqrt(10) = sqrt(90),

and

MB = 2×sqrt(10) = sqrt(40)

now, we need to calculate back using the same Pythagoras approach (calling the coordinates of M xm and ym)

AM² = (xm - -10)² + (ym - 2)² = (xm+10)² + (ym-2)² = 90

90 = xm² + 20xm + 100 + ym² - 4ym + 4

MB² = (5 - xm)² + (7 - ym)² = 40

40 = 25 - 10xm + xm² + 49 - 14ym + ym²

as a first approach we calculate AM² - MB²

90 = xm² + 20xm + ym² - 4ym + 104

- 40 = xm² - 10xm + ym² - 14ym + 74

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

50 = 0 + 30xm + 0 + 10ym + 30

20 = 30xm + 10ym

2 = 3xm + ym

ym = 2 - 3xm

this we use now e.g. in the first equation for AM.

90 = xm² + 20xm + (2-3xm)² - 4×(2-3xm) + 104

-14 = xm² + 20xm + 4 - 12xm + 9xm² - 8 + 12xm

-10 = 10xm² + 20xm

-1 = xm² + 2xm

xm² + 2xm + 1 = 0

solving such a squared equation

xm = (-b ± sqrt(b² - 4ac))/(2a)

a = 1

b = 2

c = 1

xm = (-2 ± sqrt(4 - 4))/2 = -1

only one combined solution (as a squared equation usually has 2 solutions).

ym = 2 - 3xm = 2 - 3×-1 = 2 + 3 = 5

so, M = (-1, 5)

8 0
2 years ago
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