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2 years ago

Find the midpoint of A and B where A has coordinates (2,9)

Mathematics

1 answer:

timofeeve [1]2 years ago

7 0

Answer:

$(\dfrac{9}{2}, \dfrac{13}{2})$

Step-by-step explanation:

The coordinates of A = (2,9)

The coordinates of B = (7,4)

Let M (x,y) be the mid point of A and B.

$M(x,y)=(\dfrac{x_1+x_2}{2}),(\dfrac{y_1+y_2}{2})\\\\=\dfrac{2+7}{2}, \dfrac{9+4}{2}\\\\=(\dfrac{9}{2}, \dfrac{13}{2})$

So, the coordinates of mid point of A and B is $(\dfrac{9}{2}, \dfrac{13}{2})$ .

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the area of a parallelogram shape land is on the square and length of its two adjacent sides are 25 m and 17 M find its diagonal

Alex787 [66]

Step-by-step explanation:

Draw diagonal AC

The triangle ABC has sides 17 and 25

Say AB is 17, BC is 25

Draw altitude on side BC from A , say h

h = 17 sin B

Area = 25*17 sin B = 408

sin B = 24/25

In ∆ ABC

Cos B = +- 7/25

= 625 + 289 — b^2 / 2*25*17

b^2 = 914 — 14*17 = 676

b = 26

h = 17*24/25 = 408/25 = 16.32

Draw the second diagonal BD

In ∆ BCD, draw altitude from D, say DE =h

BD^2 = h^2 + {(25 + sqrt (289 -h^2) }^2

BD^2 = 16.32^2 + (25 + 4.76)^2

= 885.6576 + 266.3424

BD = √ 1152 = 33.94 m

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2 years ago

The perimeter of a rectangular field is 314 m.

Ksivusya [100]

Answer:

85m

Step-by-step explanation:

Perimeter of a rectangle = 2(l+b)

b=72m, l=?

perimeter =314m

314=2(l+72)

314/2= l+72

157=l+72

l=157-72

l=85m

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1 year ago

Find the volume of a hemisphere whose radius is 7.5

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2 years ago

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Advocard [28]

Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer. That matters because the derivative of a constant is zero.

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$\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
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\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
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\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\
$

$\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
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\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
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\cfrac{d\theta }{dt}=\cfrac{11}{75}$

8 0

2 years ago