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2 years ago

I need help ASAP plz

Mathematics

2 answers:

Sladkaya [172]2 years ago

4 0

This is a right triangle. To find the third side of a right triangle. Use the Pythagorean Theorem.
a^2+b^2=c^2
a and b are the legs, c is the hypotenuse.
a^2+7^2=16^2
a^2+49=256
Subtract 49 from both sides:
a^2=207
Square foot:
a=14.38749457 in.
Round to nearest tenth:
14.4 in

Anit [1.1K]2 years ago

3 0

Answer:

a = 14.4

Step-by-step explanation:

It's a right angled triangle. so it has to follow Pythagoras Theorem

according to it square of hypotenuse = sum of squares of other two sides

$16 {}^{2} = {x}^{2} + 7 {}^{2} \\ 256 = x {}^{2} + 49 \\ 256 - 49 = x {}^{2} \\ 207 = x {}^{2} \\ x = \sqrt{207 } \\ x = 14.387 \\ x = 14.4$

after rounding the decimal a= 14.4 in

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How to write three hundred thousand, five thousand sixty three in standard form

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3 years ago

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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

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

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

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

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

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

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

=> 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.

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