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3 years ago

Answer this question for me please cause I'm so lost and I have no clue what I am doing

Mathematics

1 answer:

Airida [17]3 years ago

6 0

For any values less than -3, the answer is undefined, because then you have a negative value for √x+3

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Whats the answer for this one???? urgently need correct answers

kondor19780726 [428]

Answer:

Refer to the attachment

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2 years ago

Which of the following is the difference of two squares?

Blizzard [7]

Answer: C

Step-by-step explanation:

$25a^{2}=(5a)^{2}$ and $36b^{6}=(6b^{3})^{2}$

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2 years ago

Can someone tell me if this adjacent , vertical or linear

babymother [125]

Answer:

adjacent

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2 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).

5 0

3 years ago

Nadusha1986 [10]

Answer:

Step-by-step explanation:

(1,1) to (2,2) and (2,2) to (3,3) and so on is (1,1)

6 0

2 years ago