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3 years ago

Please someone help!!!

Mathematics

1 answer:

kari74 [83]3 years ago

6 0

Answer:

0.001189 x 10^7

Step-by-step explanation:

4.1 x 10^7 gallons of water

2,900 gallons of water = 2.9 x 10^3

Balance the 10th power with the first one.

2,900 = 0.00029 x 10^7

Multiply only the decimal number...

4.1 x 0.00029 = 0.001189

The answer is...

0.001189 x 10^7

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Math help :) pleasee , and thnx

mylen [45]

Answer:

The ball is in the air for approximately 3.27 seconds ⇒ answer A

Step-by-step explanation:

* Lets explain how to solve the problem

- The height of the ball is modeled by the function

h(t) = -4.9 t² + 16 t

- We need to find the time that the ball is in the air

- The ball is in the air from its initial position and then return to the

same position

- That means h(t) = 0 because h(t) represent the height of the ball

from its initial position

∵ h(t) = -4.9 t² + 16 t

∵ h(t) = 0

∴ 0 = -4.9 t² + 16 t

- Add 4.9 t² to both sides

∴ 4.9 t² = 16 t

- Subtract 16 t from both sides

∴ 4.9 t² - 16 t = 0

- Take t as a common factor

∴ t (4.9 t - 16) = 0

- Equate each factor by 0

∴ t = 0 and 4.9 t - 16 = 0

∵ 4.9 t - 16 = 0 ⇒ add 16 for both sides

∴ 4.9 t = 16

- Divide both sides by 4.9

∴ t = 3.2653

∴ t = 0 ⇒ initial position

∴ t = 3.2653 seconds ⇒ final position

* <em>The ball is in the air for approximately 3.27 seconds</em>

6 0

4 years ago

Solve the equation 2 sec² x = 3 - tan x for the domain 0° ≤ x ≤ 360°

vodka [1.7K]

$2 \sec^2 x = 3- \tan x \\\\\implies 2(1 +\tan^2 x) = 3- \tan x \\\\\implies 2 + 2 \tan^2 x +\tan x -3 =0\\\\\implies 2 \tan^2 x + \tan x -1 =0\\\\\implies 2u^2 + u -1 =0~~~ ;[\text{set} \tan x = u]\\\\\implies u = \dfrac{-1\pm \sqrt{1-4\cdot 2 \cdot (-1)}}{2(2)}\\\\\implies u =\dfrac{-1 \pm\sqrt{9}}{4}\\\\\implies u = \dfrac{-1 \pm 3}{4}\\\\\implies u = \dfrac{2}{4}=\dfrac 12~~ \text{or} ~~u =\dfrac{-4}{4} =-1\\\\$

$\implies \tan x = \dfrac 12 ~~ \text{or} ~~ \tan x =-1~~~ ;[\text{Substitute back u =tan x}]\\\\\text{Now,}~ \\\\\tan x = -1,\\\\\implies x = n\pi - \dfrac{\pi}4\\\\\\\text{For interval,}~ [0,2\pi) \\\\x = \dfrac{3\pi}4, \dfrac{7\pi}4\\\\\text{In degrees,}~ x = 135^{\circ}, x =315^{\circ}\\\\\tan x = \dfrac 12\\\\\implies x = n\pi + \tan^{-1} \left(\dfrac 12 \right)\\\\\text{For interval} ~[0,2\pi),\\\\$

$x=\tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\text{in degrees,} ~~x=\tan^{-1} \left(\dfrac 12 \right), \left[180^{\circ} +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\\\\text{Combine all solutions:}\\\\x=\dfrac{3\pi}4, \dfrac{7\pi}4, \tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\$