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3 years ago

12

What is the area of the quadrilateral with vertices at (1, 0), (2, 0), (2, 5), and (1, 5)?

2 answers:

TiliK225 [7]3 years ago

8 0

Answer:

5 unitsquare

Step-by-step explanation:

rectangle

Area: (2-1)*(5-0)=5

exis [7]3 years ago

7 0

The answer is 5 unit square

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The two non-parallel sides of an isosceles trapezoid are each 7 feet long. The longer of the two bases measures 22 feet long. Th

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The length of the diagonal and the shorter base are 17. 24 feet and 18. 7 feet long respectively.

<h3>How to determine the length</h3>

The cosine rule is given as;

$c = \sqrt{a^2 + b^2 - 2ab cos \alpha }$

c = length of the diagonal

a = base length = 22 feet

b = 7 feet

$c = \sqrt{7^2 + 22^2 - 2 * 7 * 22 cos 40}$

$c = \sqrt{533 - 308 * 0. 7660}$

$c = \sqrt{533 - 235. 93}$

$c = \sqrt{297. 072}$

$c = 17. 24$ feet

Using sine rule

$\frac{a}{sin A } = \frac{c}{sin C}$

$\frac{a}{sin 70} = \frac{17. 24}{sin 40}$

Cross multiply

$a$ × $sin 60$ = $c$ × $sin 70$

$a$ × $0. 8660$ = $17. 24$ × $0. 9397$

$a = \frac{16. 20}{0. 8660}$

a = 18. 7 feet long

Thus, the length of the diagonal and the shorter base are 17. 24 feet and 18. 7 feet long respectively.

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Marina86 [1]

Answer:

Equation of Sphere = $x^{2}$ + $y^{2}$ + $z^{2}$ - 18x - 4/3y +10z + 142/3 = 0

Step-by-step explanation:

Data Given:

P = (x,y,z)

Distance from P to A (-3,6,3) = Twice the distance from P to B(6,2,-3)

Solution:

Find the equation of the sphere:

It is given that:

PA = 2PB

Squaring both sides:

$(PA)^{2} = (2PB)^{2}$

$(PA)^{2} = 4 (PB)^{2}$

$(x - (-3))^{2}$ + $(y - 6)^{2}$ + $(z-3)^{2}$ = 4 x $(x-6)^{2}$ + $(y-2)^{2}$ + $(z-(-3))^{2}$

Solving the above equation:

$(x + 3))^{2}$ + $(y - 6)^{2}$ + $(z-3)^{2}$ = 4 x { $(x-6)^{2}$ + $(y-2)^{2}$ + $(z + 3))^{2}$ }

$x^{2}$ + 9 + 6x + $y^{2}$ + 36 - 12y + $z^{2}$ + 9 - 6z = 4 { $x^{2}$ + 36 - 12x + $y^{2}$ + 4 - 4y + $z^{2}$ + 9 + 6z}

$x^{2}$ + 9 + 6x + $y^{2}$ + 36 - 12y + $z^{2}$ + 9 - 6z = 4 $x^{2}$ + 144 - 48x +4 $y^{2}$ + 16 - 16y + 4 $z^{2}$ + 36 + 24z

Putting the right hand side = 0 and solving the equation:

$x^{2}$ - 4 $x^{2}$ + $y^{2}$ - 4 $y^{2}$ + $z^{2}$ - 4 $z^{2}$ + 6x + 48x - 12y +16y -6z - 24z + 9 + 36 + 9 -144 - 16 - 36 = 0

-3 $x^{2}$ - 3 $y^{2}$ -3 $z^{2}$ + 54x + 4y -30z -142 = 0

Taking (-) common

- ( 3 $x^{2}$ + 3 $y^{2}$ + 3 $z^{2}$ - 54x - 4y +30z + 142) = 0

3 $x^{2}$ + 3 $y^{2}$ + 3 $z^{2}$ - 54x - 4y +30z + 142 = 0

dividing the whole equation by 3

$x^{2}$ + $y^{2}$ + $z^{2}$ - 54/3x - 4/3y +30/3z + 142/3 = 0

$x^{2}$ + $y^{2}$ + $z^{2}$ - 54/3x - 4/3y +30/3z + 142/3 = 0

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3 years ago