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Taya2010 [7]
3 years ago
15

4. A rectangle has a diagonal of 8 cm. The diagonal creates a 60° angle at the base of the rectangle.

Mathematics
1 answer:
tamaranim1 [39]3 years ago
7 0

9514 1404 393

Answer:

  a) see attachment

  b) base: 4, height: 4√3

  c) area: 16√3

Step-by-step explanation:

a) The figure is drawn in the attachment

__

b) The side lengths of a 30-60-90 right triangle have the ratio ...

  1 : √3 : 2

Multiplying these by 4 gives the side lengths of each triangle created by the diagonal. The base (short leg) is 4 units; the height (long leg) is 4√3 units.

__

c) The area is the product of base and height:

  A = bh

  A = 4(4√3) = 16√3 . . . . square units

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Solve the equation using the zero-product property (2x-8)(7x+5)=0
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Answer:

\boxed{2^{\frac{802}{27}} \cdot 3^9}

Step-by-step explanation:

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$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

\left(\dfrac{1}{3^2}  \cdot \dfrac{1}{4^5}  \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3

Also,

\boxed{a^{0} = 1, a\neq 0 }

\left(\dfrac{1}{9}  \cdot \dfrac{1}{1024}  \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

Note

\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq  0}

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27

\left(\dfrac{1}{9216}   \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)

Note

\boxed{\dfrac{1}{\dfrac{1}{a} }  = a}

9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)

\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)

Once

9216=2^{10}\cdot 3^2 \implies  9216^3=2^{30}\cdot 3^6

\boxed{(a \cdot b)^n=a^n \cdot b^n}

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27} $

We have

\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)

Also, once

\boxed{\dfrac{c^a}{c^b}=c^{a-b}}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27

As

30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27}  =\dfrac{802}{27}

2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3

2^{\frac{802}{27}} \cdot 3^9

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