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sergey [27]
2 years ago
11

Question 10 The hypotenuse of a right triangle is I m longer than the longer leg. The other leg is 7 m shorter than the longer l

eg. Determine the lengths of the three sides of the triangle. (3 marks)​
Mathematics
1 answer:
bonufazy [111]2 years ago
5 0

Answer:

5, 12, 13

Step-by-step explanation:

let x be the longer leg then x + 1 is the hypotenuse and x - 7 the shorter leg

Using Pythagoras' identity in the right triangle

x² + (x - 7)² = (x + 1)² ← expand using FOIL

x² + x² - 14x + 49 = x² + 2x + 1

2x² - 14x + 49 = x² + 2x + 1 ( subtract x² + 2x + 1 from both sides )

x² - 16x + 48 = 0 ← in standard form

(x - 4)(x - 12) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 4 = 0 ⇒ x = 4

x - 12 = 0 ⇒ x = 12

x = 4 , then x - 7 = 4 - 7 = - 3 ← not possible

x = 12, then x - 7 = 12 - 7 = 5 and x + 1 = 12 + 1 = 13

The lengths of the 3 sides are

longer leg = 12 m , shorter leg = 5 m and hypotenuse = 13 m

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3 years ago
If x+y=6 and xy =2find x³+y³(please help me fast with explanation also please) T^T​
Maurinko [17]

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Step-by-step explain:                                                                                            let's remember the formula                                                                                 x³+y³=(x+y)(x²-xy+y²) and also  x³+y³=(x+y)³-3xy(x+y)                                              then                                                                                      \displaystyle\boldsymbol{ x^3+y^3=\underbrace{(x+y)^3}_{6}-3\underbrace{xy}_{2}\underbrace{(x+y)}_6}=\\\\6^3-3\cdot 2\cdot 6=216-36=180                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                

6 0
3 years ago
Given f(x) =
sergejj [24]

Answer:

A

Step-by-step explanation:

We are given the function:

\displaystyle f(x) = \left\{        \begin{array}{ll}            2\cos(\pi x) \text{ for }  x \leq -1 \\ \\          \displaystyle   \frac{2}{\cos(\pi x)}\text{ for } x > -1        \end{array}    \right.

And we want to find:

\displaystyle \lim_{x\to -1}f(x)

So, we need to determine whether or not the limit exists. In other words, we will find the two one-sided limits.

Left-Hand Limit:

\displaystyle \lim_{x\to-1^-}f(x)

Since we are approaching from the left, we will use the first equation:

\displaystyle =\lim_{x\to -1^-}2\cos(\pi x)

By direct substitution:

=2\cos(\pi (-1))=2\cos(-\pi)=2(-1)=-2

Right-Hand Limit:

\displaystyle \lim_{x\to -1^+}f(x)

Since we are approaching from the right, we will use the second equation:

=\displaystyle \lim_{x\to -1^+}\frac{2}{\cos(\pi x)}

Direct substitution:

\displaystyle =\frac{2}{\cos(\pi (-1))}=\frac{2}{\cos(-\pi)}=\frac{2}{(-1)}=-2

So, we can see that:

\displaystyle \displaystyle \lim_{x\to-1^-}f(x)=\displaystyle \lim_{x\to -1^+}f(x) =-2

Since both the left- and right-hand limits exist and equal the same thing, we can conclude that:

\displaystyle \lim_{x \to -1}f(x)=-2

Our answer is A.

8 0
3 years ago
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