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2 years ago

Use the distributive property to remove the parentheses. -3(-6y +3w-2)

Mathematics

2 answers:

wariber [46]2 years ago

5 0

Answer:

18y-9w+6

Step-by-step explanation:

The distributive property tells that because the -3 is outside of the parentheses, we multiply all values inside the parentheses by -3.

So, -6y * -3 is 18y because two negatives cancel out, 3w * -3 equals -9w, and -2 * -3 is 6, again because two negatives cancel each other out.

We put this all together, the parentheses and -3 is removed, and the expression becomes 18y-9w+6.

LekaFEV [45]2 years ago

3 0

Answer:

18y-9w+6

Step-by-step explanation:

-3*-6y=18y

-3*3w=-9w

-3-2=6

so, 18y-9w+6

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The ellipse with x-intercepts (2, 0) and (-2, 0); y-intercepts (0, 4) and (0, -4).

masya89 [10]

Answer:

$\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$

Step-by-step explanation:

If the ellipse has its x-intercepts at points (2, 0) and (-2, 0) and y-intercepts at points (0, 4) and (0, -4), then its symmetric across the y-axis and across the x-axis.

Moreover,

$a=2\\ \\b=4$

The equation of such ellipse is

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

Hence, the equation of the ellipse is

$\dfrac{x^2}{2^2}+\dfrac{y^2}{4^2}=1\\ \\ \\\dfrac{x^2}{4}+\dfrac{y^2}{16}=1$

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.