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ipn [44]
3 years ago
13

I NEED HELP ASAPPPPPPPPPPPP

Mathematics
2 answers:
Alekssandra [29.7K]3 years ago
7 0

The answer to your question is C

Bingel [31]3 years ago
4 0

Answer:

c.

Step-by-step explanation:

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2 years ago
Pedro is building a playground in the shape of a right triangle he wants to know the area of the playground to help him decide h
shusha [124]

Answer:

Amount of sand=Area of triangle(ABC)=1/2*AB*BC

Dimension of rectangle as  length=BC and breadth=AB

Step-by-step explanation:

Given:

Pedro building a playground in shape of right angled triangle.

To Find:

How much sand he need to buy

And if playground changed to rectangle what will be the dimensions.

Solution:

<em>Consider a ΔABC be the play ground vertex of playground,</em>

<em>And AB and BC be the sides making right angle.</em>

The sand required to fill ground will be the amount of are covered by the ABC triangle.

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Area Of triangle(ABC)=1/2* base* height

Here base will be BC and height =AB

Therefore Area of Triangle(ABC)=1/2*BC*AB

Depending on the lengths of base and height amount of sand will be decided.

Now,

<em>For Rectangle dimensions,</em>

<em>We know that if same sized triangles composes each other forms a rectangle</em>.

It requires two triangle to form one rectangle as follows

(Refer the attachment)

Same sized Triangle ACD is imposed on it  to from rectangle ABCD.

So dimension for rectangle will be same as the triangle

Dimension=length and breadth

i.e. length=base of triangle=BC

Breadth=Height of triangle=AB

i.e Breadth =AB.

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3 years ago
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4 years ago
Solve the system by substitution.<br> - 8x + y = -32<br> - 3x – 10 = y<br> (,)
agasfer [191]

Answer:

x=2,\:y=-16

Step-by-step explanation:

\begin{bmatrix}-8x+y=-32\\ -3x-10=y\end{bmatrix}\\\\\mathrm{Isolate}\:x\:\mathrm{for}\:-8x+y=-32:\quad x=-\frac{-32-y}{8}\\\\\mathrm{Subsititute\:}x=-\frac{-32-y}{8}\\\begin{bmatrix}-3\left(-\frac{-32-y}{8}\right)-10=y\end{bmatrix}\\\\Simplify\\\begin{bmatrix}\frac{3\left(-32-y\right)}{8}-10=y\end{bmatrix}\\\\\mathrm{Isolate}\:y\:\mathrm{for}\:\frac{3\left(-32-y\right)}{8}-10=y:\quad y=-16\\\\\mathrm{For\:}x=-\frac{-32-y}{8}\\\\\mathrm{Subsititute\:}y=-16\\x=-\frac{-32-\left(-16\right)}{8}\\

-\frac{-32-\left(-16\right)}{8}=2\\x=2\\x=2,\:y=-16

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4 years ago
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