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Alecsey [184]
2 years ago
14

??????????????????????????????????????????????

Mathematics
1 answer:
Mars2501 [29]2 years ago
3 0
The first one is 2x^2+2x-2
the second one is 2x^2-2x-2
the third one is 6h+6k
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The perimeter of a rectangle is 30.8 km and it’s diagonal length is 11 km. Find it’s length and width
blsea [12.9K]

Answer:

Length of the rectangle is 15.0325 km and width is 0.3765 km.

Explanation:

Given:

Perimeter of a rectangle = 30.8 km

Length of diagonal of rectangle = 11 km

To find:

The length and width of rectangle=?

Solution:

Lets assume length of the rectangle = x km

And assume width of the rectangle = y km

Lets first create equation using given  perimeter

perimeter of rectangle = 2 ( length +  width )

=> 30.8 km = 2 ( x + y )  

=>x + y = \frac{30.8}{2}

=> y = 15.4 – x             ------(1)

As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,  

=> length^2 + width^2 = diagonal^2

=> x^2 + y^2 = 11^2

=> x^2 + y^2 = 121

On substituting value of y from (1) in above equation we get

=> x^2 + (15.4-x)^2 = 121

=>x^2 + (15.4)^2 + x^2 – 2 x 15.4 \times x   = 121

=> 2x^2-30.8x + 237.16 -121  = 0

=> 2x^2-30.8x + 116.16 = 0

Solving above quadratic equation using quadratic formula

General form of quadratic equation is  

ax^2 +bx +c = 0

And quadratic formula for getting roots of quadratic equation is  

x= \frac{ -b\pm\sqrt{(b^2-4ac)}}{2a}

As equation is 2x^2-30.8x + 116.16 = 0, in our case

a = 2 ,  b = -30.8 and c = 116.16

Calculating roots of the equation we get

x=\frac{ -(-30.8)\pm\sqrt{(-30.8)^2-4(2)( 11)} } {(2\times2)}

x=\frac{30.8\pm\sqrt{(948.64-88)}}{4}

x=\frac{30.8\pm\sqrt{860.64}}{4}

x=\frac{30.8\pm\sqrt{860.64}}{4}

x=\frac{(30.8\pm29.33)}4

x=\frac{(30.8+29.33)}{4}

x=\frac{(30.8-29.33)}{4}

=> x = 15.0325 or x = 0.3675

As generally length is longer one ,  

So x = 1.0325

From equation (1) y = 15.4 – x = 0.3765

Hence length of the rectangle is 15.0325 km and width is 0.3765 km.

6 0
3 years ago
(x-3)^2=5 ? confused
Vanyuwa [196]

(x-3)^2=5\implies \sqrt{(x-3)^2}=\sqrt{5}\implies (x-3)=\sqrt{5}\implies x = \sqrt{5}+3

5 0
2 years ago
Find the distance between points A (5.-7) and B (-3,-4)
Strike441 [17]

It would help if you had a four quadrant graph, but the distance would be (8,11).

7 0
3 years ago
Which graph represents the function f(x)=(x-5)^2+3
hammer [34]

Answer:

Parabola

Step-by-step explanation:

We are given that a function

f(x)=(x-5)^2+3

The given function is an equation of parabola along y- axis.

General equation of parabola along y- axis  with vertex (h,k) is given by

(y-k)=(x-h)^2

y=(x-h)^2+k

Compare it with given equation then we get

h=5, k=3

Vertex of given parabola =(5,3)

Substitute x=0 then we get

f(0)=(0-5)^2+3=28

y-intercept of parabola is at (0,28).

6 0
3 years ago
Read 2 more answers
1. List all the possible rational solutions for each equation.
12345 [234]

Answer:

x=−7.974656,0.038604

Step-by-step explanation:

4x4+13x3−124x2+212x−8=0

Brainleist please

5 0
2 years ago
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