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1 year ago

Calculate the length of AC to one decimal place in the trapezium below.

Mathematics

1 answer:

a_sh-v [17]1 year ago

7 0

Answer:

First we will split trapezium on two geometric figure.

One is rectangle and the second is right triangle.

When we subtract 11-4=7cm we get one cathetus of the right triangle a=7cm also we know hypotenuse which is c=16cm.

We will use Pythagorean theorem to find the second cathetus b

b=√16∧2-7∧2= √256-49= √207 ≈ 14.39cm =DC =>

AC =√(DC)∧2+(AD)∧2= √14.39∧2+11∧2= √207+121= √328= 18.1cm

AC=18.1cm

Good luck!!

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If the limit that you want to find is $\lim_{x\to \infty}\dfrac{P(x)}{Q(x)}$ then you can use the following proof.

Step-by-step explanation:

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ and $Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0}$ be the given polinomials. Then

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and

$\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}$

Then

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