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3 years ago

21 p?=u² + 2as u= 12 a = -3 s= 18 (a) Work out a value of v.

Mathematics

1 answer:

aalyn [17]3 years ago

6 0

Answer:

6 or -6

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

$v^2 = 12^2 + 2(-3)(18)$
$v^2 = 144 + (-6)(18)$
$v^2 = 144- 108$
$v^2 = 36$

Step 2: Take square root of both sides.

$\sqrt{v^2} =$ ± $\sqrt{36}$
$v =$ ± $\sqrt{36}$
$v = 6$ or $-6$

Therefore, the answer is 6 or -6.

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The average price of a movie ticket in the United States from 1980 to 2000 can be modeled by the expression, 2.75+0.10x, where x

MissTica

Answer:

Step-by-step explanation:

step one:

Let the cost be y

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3 years ago

A surveyor leaves her base camp and drives 42km on a bearing of 032degree she then drives 28km on a bearing of 154degree,how far

ValentinkaMS [17]

Answer:

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

Step-by-step explanation:

The final position of the surveyor is represented by the following vectorial sum:

$\vec r = \vec r_{1} + \vec r_{2} + \vec r_{3}$ (1)

And this formula is expanded by definition of vectors in rectangular and polar form:

$(x,y) = r_{1}\cdot (\cos \theta_{1}, \sin \theta_{1}) + r_{2}\cdot (\cos \theta_{2}, \sin \theta_{2})$ (1b)

Where:

$x, y$ - Resulting coordinates of the final position of the surveyor with respect to origin, in kilometers.

$r_{1}, r_{2}$ - Length of each vector, in kilometers.

$\theta_{1}, \theta_{2}$ - Bearing of each vector in standard position, in sexagesimal degrees.

If we know that $r_{1} = 42\,km$ , $r_{2} = 28\,km$ , $\theta_{1} = 32^{\circ}$ and $\theta_{2} = 154^{\circ}$ , then the resulting coordinates of the final position of the surveyor is:

$(x,y) = (42\,km)\cdot (\cos 32^{\circ}, \sin 32^{\circ}) + (28\,km)\cdot (\cos 154^{\circ}, \sin 154^{\circ})$

$(x,y) = (35.618, 22.257) + (-25.166, 12.274)\,[km]$

$(x,y) = (10.452, 34.531)\,[km]$

According to this, the resulting vector is locating in the first quadrant. The bearing of the vector is determined by the following definition:

$\theta = \tan^{-1} \frac{10.452\,km}{34.531\,km}$

$\theta \approx 16.840^{\circ}$

And the distance from the camp is calculated by the Pythagorean Theorem:

$r = \sqrt{(10.452\,km)^{2}+(34.531\,km)^{2}}$

$r = 36.078\,km$

The surveyor is 36.076 kilometers far from her camp and her bearing is 16.840° (standard form).

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