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alexgriva [62]
2 years ago
10

Please help me with this Evaluate: 2^-4 =

Mathematics
1 answer:
givi [52]2 years ago
5 0

Answer:  \frac{1}{16}

Step-by-step explanation:

Whenever an exponent is negative, you flip the solution.

For this, 2^4 = 2*2*2*2 = 4*4 = 16, however the exponent is negative, so flip 16 to get your final answer of \frac{1}{16}

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In a trapezoid the lengths of bases are 11 and 18. The lengths of legs are 3 and 7. The extensions of the legs meet at some poin
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Answer:  7\frac{5}{7} unit and 18 unit

Step-by-step explanation:

Let ABCD is a trapezoid where AB and CD are the bases. ( In which AB is greatest base which shown in below figure)

AD and BC are the legs of the trapezoid ABCD.

Now, we have ( According to the question ),

AB = 18 unit, BC = 7 unit, AD = 3 unit and DC = 11 unit.

Here the leg AD extends from point D.

Similarly leg BC extends from point C.

Let they meet at point P ( shown in below diagram)

Since In triangles PAB and PDC,

∠PDC ≅ ∠PAB ( because DC ║ AB )

And, ∠ PAB ≅ ∠ PBA

∠DPC ≅ ∠ APB ( reflexive)

Therefore, By AAA similarity postulate,

\triangle PDC \sim \triangle PAB

Thus, By the definition of similarity,

\frac{PD}{PA} = \frac{DC}{AB}

\frac{PD}{PD+3} = \frac{11}{18} ( because PA = PD+DA)

⇒ 18 PD = 11 PD +33

⇒7PD = 33

⇒ PD = 33/7

Again by the definition of similarity,

\frac{PC}{PB} = \frac{DC}{AB}

\frac{PC}{PC+7} = \frac{11}{18} ( because PB = PC + CB)

⇒ 18 PC = 11 PD +77

⇒7PC = 77

⇒ PC = 11

Thus, PA =  PD+DA = 33/7 + 3 = 7\frac{5}{7}

And, PB = PC + CB = 11 + 7 = 18


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