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3 years ago

What is the perimeter of a rectangle with sides 2/3x+10 and 1/3x+5

Mathematics

1 answer:

gogolik [260]3 years ago

8 0

Answer:

The perimeter of the rectangle is 2x+ 30

Step-by-step explanation:

Sides of the rectangle are 2/3x+10 and 1/3x+5

Now, Perimeter of the Rectangle = 2(Length + Breadth)

Here, sum of the sides = $(\frac{2}{3}x + 10 ) + (\frac{1}{3}x + 5 ) = (\frac{2}{3}x +\frac{1}{3}x ) + ( 10+ 5 )$

adding like terms with like terms, we get

( $\frac{2x + x}{3}) + 15 = \frac{3x}{3} + 15$

Hence, Sum = (x + 15)

Now, 2(Length + Breadth) = 2(x+15) = 2x + 30

hence, the perimeter of the rectangle = 2x+ 30

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<h2>1. y-intercept</h2>

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$f(x)=2x^2-8x-10 \\ \\ If \ x=0 \rightarrow f(0)=-10 \\ \\ Then \ y-intercept: \\ \\ (0,-10)$

<h2>2. x-intercepts</h2>

$\boxed{(-1,0) \ and \ (5,0)}$

To find the other x-intercept, we must set $y=0$ as follows:

$f(x)=2x^2-8x-10 \\ \\ If \ y=0 \rightarrow 2x^2-8x-10=0 \\ \\ Using \ the \ quadratic \ formula: \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \therefore x=\frac{-(-8) \pm \sqrt{(-b)^2-4(2)(-10)}}{2(2)} \\ \\ \therefore x_{1}=-1 \ and \ x_{2}=5$

Therefore, the other x-intercept is $(5,0)$ . You can see both the y-intercept and the x-intercepts in the figure below.

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