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2 years ago

Examine the equation.

Mathematics

2 answers:

kow [346]2 years ago

7 0

Infinity many solutions

dezoksy [38]2 years ago

4 0

Infinity many solutions

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Given that a function, g, has a domain of -20 sxs 5 and a range of 5 s 39 s 45 and that g(0) = -2 and g(-9) = 6, select the stat

luda_lava [24]

Answer:

option (B) and (D)

Step-by-step explanation:

It's just basic of domain and range just check out the domain and range you can get your answer .

Only option (B) and (D) satisfy in the given range and domain

8 0

3 years ago

To the nearest tenth, which is the perimeter of ABC. Geometry

qwelly [4]

Answer:

23.6

Step-by-step explanation:

Finding AC:

Cos 61 = $\frac{adjacent}{hypotenuse}$

0.48 × 10 = Adjacent

AC = 4.8

Now, CB:

Cos 29 = $\frac{adjacent}{hypotenuse}$

0.87 × 10 = CB

CB = 8.8

The perimeter:

=> 10+4.8+8.8

=> 23.6

3 0

2 years ago

Read 2 more answers

Please help with this!!

maw [93]

Answer:

gcf=2

12-2=4(3-1)

hope this is right!

8 0

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

Alex types 1/2 of a page in 1/2 an hour. How long will it take him to type 1 page?

muminat

B. 1 hour, he takes the exact amount of time that he types.

5 0

3 years ago