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adell [148]
3 years ago
10

A rectangular hockey field is 50m long and 35m wide calculate the length of it's diagonal​

Mathematics
1 answer:
Pavlova-9 [17]3 years ago
3 0

Answer:

hope it helps u

plz mark brainliest✌️✌️

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Help please will give brainly if <br> answerd
Fynjy0 [20]

Answer:

its the middle one

Step-by-step explanation:

had this on a test, got it right

7 0
3 years ago
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Find a transformation of sin(x) that gives the following <br><br> a. <br><br> b.
riadik2000 [5.3K]

Answer:

a) f(x) = 3sin(2x - π/2) + 1

b) f(x) = 2sin(2x + π) + 2

Step-by-step explanation:

∵ f(x) = Asin(Bx - C) + D

Where ΙAΙ is the the amplitude , 2π/ΙBΙ is the period , -C/B is the horizontal shift and D is the vertical shift

a) The amplitude is 3

   The period is π

    Shift phase: horizontally π/4 to the right , vertically 1 unit up

b) The amplitude is 2

   The period is π

    Shift phase: horizontally π/2 to the left , vertically 2 unit up

4 0
3 years ago
For the upcoming holiday season, Dorothy wants to mold 20 bars of chocolate into tiny pyramids. Each bar of chocolate contains 6
gizmo_the_mogwai [7]
So each bar contains 6 and she has 20 so 6 times 20=120

pyramid volume=legnth times width times height times 1/3
v=base times height times 1/2
v=1 times 2 times 1/3
v=2/3
volume of 1 pyramid is 2/3 cubic inches

divide 120 by 2/3 to find out how many pyrimds will fit into amount given
120/1 divide by 2/3=120/1 times 3/2=360/2=180

she can make 180 pyramids
3 0
3 years ago
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Among all right triangles whose hypotenuse has a length of 12 cm, what is the largest possible perimeter?
Veronika [31]

Answer:

Largest perimeter of the triangle =  

P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)

Step-by-step explanation:

We are given the following information in the question:

Right triangles whose hypotenuse has a length of 12 cm.

Let x and y be the other two sides of the triangle.

Then, by Pythagoras theorem:

x^2 + y^2 = (12)^2 = 144\\y^2 = 144-x^2\\y = \sqrt{144-x^2}

Perimeter of Triangle = Side 1 + Side 2 + Hypotenuse.

P(x) = x + \sqrt{144-x^2} + 12

where P(x) is a function of the perimeter of the triangle.

First, we differentiate P(x) with respect to x, to get,

\frac{d(P(x))}{dx} = \frac{d(x + \sqrt{144-x^2} + 12)}{dx} = 1-\displaystyle\frac{x}{\sqrt{144-x^2}}

Equating the first derivative to zero, we get,

\frac{dP(x))}{dx} = 0\\\\1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0

Solving, we get,

1-\displaystyle\frac{x}{\sqrt{144-x^2}} = 0\\\\x = \sqrt{144-x^2}}\\\\x^2 = 144-x^2\\\\x = \sqrt{72} = 6\sqrt{2}

Again differentiation P(x), with respect to x, using the quotient rule of differentiation.

\frac{d^2(P(x))}{dx^2} = \displaystyle\frac{-(144-x^2)^{\frac{3}{2}}-x^2}{(144-x)^{\frac{3}{2}}}

At x = 6\sqrt{2},

\frac{d^2(V(x))}{dx^2} < 0

Then, by double derivative test, the maxima occurs at x = 6\sqrt{2}

Thus, maxima occurs at x = 6\sqrt{2} for P(x).

Thus, largest perimeter of the triangle =  

P(6\sqrt{2}) = 6\sqrt{2} + \sqrt{144-72} + 12 = 12\sqrt{2} + 12 = 12(\sqrt2 + 1)

7 0
3 years ago
I need help ASAP marking brainliest
Thepotemich [5.8K]

Answer: option D -1,0,2

Step-by-step explanation:

2^x-2

When x=0

We have 2^0-2

1-2=-1

when x=1

We have 2^1-2

2-2=0

When x=2

We have 2^2-2

4-2=2

6 0
3 years ago
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