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SOVA2 [1]
3 years ago
11

8.9 x 10^-10 in standard form

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
6 0

Answer:

science E-notation: 8.900e-10

Engineering Notation: 890.0x-12

real number: 8.9E-10

-.-

Step-by-step explanation:

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Answer:

x = 15°

Step-by-step explanation:

These angles form a linear

4x - 17 + 137° = 180°

4x - 17 = 43°

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x = 15°

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Projectile Is shot upward from the surface of Earth with an initial velocity of 136 meters per second. Use the position
Anna007 [38]

Answer:

v(3\,s) = 106.6\,\frac{ft}{s}, v(15\,s) = -11\,\frac{ft}{s}

Step-by-step explanation:

The velocity function of the projectile is obtained by deriving the position function:

v(t) = -9.8\cdot t + 136

Now, the velocities after 3 and 15 seconds are, respectively:

v(3\,s) = -9.8\cdot (3) + 136

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v(15\,s) = -9.8\cdot (15) + 136

v(15\,s) = -11\,\frac{ft}{s}

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3 years ago
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.

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