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2 years ago

Sin^2x + sin2x +2cox^2x = 2

Mathematics

1 answer:

Effectus [21]2 years ago

5 0

Answer:

Step-by-step explanation:

$sin^2 \ x + sin ^2 \ x + 2cos^ 2\ x = 2\\\\2sin^2 \ x + 2 cos^2 \ x = 2\\\\2 ( sin^2 \ x + cos^2 \ x ) = 2 \\\\2 \times 1 = 2 \\\\2 = 2 \\\\LHS = RHS \\\\Hence \ proved.$

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Simplify the expression.<br> 7x2 + 3 - 5(x^2- 4)

polet [3.4K]

Answer:

2x² + 23

Step-by-step explanation:

Distribute the -5 and combine like terms.

7x² + 3 - 5(x² - 4) =

= 7x² + 3 - 5x² + 20

= 2x² + 23

3 0

3 years ago

A ball is thrown vertically in the air with a velocity of 95 ft/s. The ball is at a height of 120 ft.

sattari [20]

Answer:

The ball is at a height of 120 feet after 1.8 and 4.1 seconds.

Step-by-step explanation:

The statement is incomplete. The complete statement is perhaps the following "A ball is thrown vertically in the air with a velocity of 95 ft/s. What time in seconds is the ball at a height of 120ft. Round to the nearest tenth of a second."

Since the ball is launched upwards, gravity decelerates it up to rest and moves downwards. The position of the ball can be determined as a function of time by using this expression:

$y = y_{o} + v_{o}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$y_{o}$ - Initial height of the ball, measured in feet.

$v_{o}$ - Initial speed of the ball, measured in feet per second.

$g$ - Gravitational constant, equal to $-32.174\,\frac{ft}{s^{2}}$ .

$t$ - Time, measured in seconds.

Given that $y_{o} = 0\,ft$ , $v_{o} = 95\,\frac{ft}{s}$ , $g = -32.174\,\frac{ft}{s^{2}}$ and $y = 120\,ft$ , the following second-order polynomial is found:

$120\,ft = 0\,ft + \left(95\,\frac{ft}{s} \right)\cdot t +\frac{1}{2}\cdot \left(-32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}$

$-16.087\cdot t^{2} + 95\cdot t -120 =0$

The roots of this polynomial are, respectively:

$t_{1} \approx 4.075\,s$ and $t_{2} \approx 1.831\,s$ .

Both roots solutions are physically reasonable, since $t_{1}$ represents the instant when the ball reaches a height of 120 ft before reaching maximum height, whereas $t_{2}$ represents the instant when the ball the same height after reaching maximum height.

In nutshell, the ball is at a height of 120 feet after 1.8 and 4.1 seconds.

8 0

3 years ago

Which is the best estimate for the average rate of change for the quadratic function graph on the interval 0 ≤ x ≤ 4?

den301095 [7]

ANSWER

A) -1

EXPLANATION

The average rate of change of the given quadratic function on the interval 0 ≤ x ≤4 is the slope of the secant line connecting the points (0,f(0)) and (4,f(4)).

That is the average rate of change is:

$= \frac{f(4) - f(0)}{4 - 0}$

From the graph, f(0) is 0 and f(4) is -4.

We plug in these values to obtain;

$= \frac{ - 4 - 0}{4 - 0}$

This simplifies to;

$= \frac{ - 4}{4}$

$= - 1$

Hence the average rate of change for the given quadratic function whose graph is shown on 0≤x≤4 is -1

5 0

3 years ago

Read 2 more answers

Naoya read a book cover to cover in a single session, at a rate of 555555 pages per hour. After 444 hours, he had 350350350 page

natali 33 [55]

Answer:

$y=-55x+570$