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erica [24]
3 years ago
5

I need help on the two bottom triangles please

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
5 0

Answer: 6. x=29yd 7. x-6.5cm


Step-by-step explanation:

6. All the sides of the square are equal (we know this because of the little lines on the sides), and the equation for finding the hypotanuse of a triangle is a^2+b^2=c^2. That's 20^2+21^2=x^2. That simplifies to 841^2=x^2, which then simplifies to x=29.


7. The bottom of the triangles are the same length, so each one is 2.5cm long. The angles are also the same, which means that the triangles are equal, so we have 2 similarities between the triangles, so they are both the same (You'll learn why later in math)

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Factorise 125x^3 -27y^3
telo118 [61]

Answer:

(5x - 3y)(25x² +15xy + 9y²)

Step-by-step explanation:

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Given

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= (5x - 3y)((5x)² + (5x)(3y) + (3y)²)

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Kim is x years old. Jordan is 9 years older than Kim. Five times Jordan’s age is equal to 110.
soldi70 [24.7K]

J=9+x

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Suppose that F is an inverse square force field below, where c is a constant.
Andre45 [30]

Answer:

Forces in our Universe

Step-by-step explanation:

a)

First of all we have,

F(r) = \frac{cr}{|r|^{3} }

and,

r = xi+yj+zk

We need to define a function that allows us to find said change based on r, so one of the functions that shows that change is,

f(r) = - c /|r|

That is,

\nabla f = F

For this case F is a conservative field and the line integral is independent of the path. Thus, defining  P_{1} = (x_{1}, y_{1}, z_{1}) and P_{2} = (x_{2}, y_{2}, z_{2}) . So the amount of work on the movement of the object from P1 to P2 is,

W=\int\limits_c  {F} \, dr

W= f(P_{1}-P_{2})

W= \frac{c}{(x_{2}^2+ y_{2}^2+ z_{2}^2)^{1/2} } -\frac{c}{(x_{1}^2+ y_{1}^2+ z_{1}^2)^{1/2}}

W= c(\frac{1}{d1}-\frac{1}{d2}  )

2) The gravitational force field is given by,

F(r) =-\frac{mMGr}{ |r|^3}

The maximum distance from the earth to the sun is 1.52*10 ^ 8 km and the minimum distance is 1.47*10 ^ 8km. The mass values of the bodies are given by m = 5.97*10 ^ {24}kg, M = 1.99 *19 ^ {30}kg and the constant G is 6.67 * 10 ^{ -11 } \frac{Nm ^ 2}{kg^2}

In this way we raise the problem like this,

c= -mMG

W= -mMG (\frac{1}{1.52*10 ^ 8} -\frac{1}{1.47*10 ^ 8} )

W= -(5.95*10^{24})(1.99*10^{30})(6.67*10^{-11})(-2.2377*10^{-10})

W \approx 1.77*10^{35}}J

5 0
3 years ago
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