Which of the following situations results in a sum of 1 1/2 ? Select all that apply .

What is the slope of the line?

Alja [10]

Answer:

4

Step-by-step explanation:

slope, m = (y₂ - y₁)/(x₂ - x₁)

plug in the values, and we get m = (3 - (-1))/((3 - 2) = 4

As x increases by 1, y increases by 4.

7 0

3 years ago

Read 2 more answers

There are 3 islands A,B,C. Island B is east of island A, 8 miles away. Island C is northeast of A, 5 miles away and northwest of

Nostrana [21]

Answer:

The bearing needed to navigate from island B to island C is approximately 38.213º.

Step-by-step explanation:

The geometrical diagram representing the statement is introduced below as attachment, and from Trigonometry we determine that bearing needed to navigate from island B to C by the Cosine Law:

$AC^{2} = AB^{2}+BC^{2}-2\cdot AB\cdot BC\cdot \cos \theta$ (1)

Where:

$AC$ - The distance from A to C, measured in miles.

$AB$ - The distance from A to B, measured in miles.

$BC$ - The distance from B to C, measured in miles.

$\theta$ - Bearing from island B to island C, measured in sexagesimal degrees.

Then, we clear the bearing angle within the equation:

$AC^{2}-AB^{2}-BC^{2}=-2\cdot AB\cdot BC\cdot \cos \theta$

$\cos \theta = \frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC}$

$\theta = \cos^{-1}\left(\frac{BC^{2}+AB^{2}-AC^{2}}{2\cdot AB\cdot BC} \right)$ (2)

If we know that $BC = 7\,mi$ , $AB = 8\,mi$ , $AC = 5\,mi$ , then the bearing from island B to island C:

$\theta = \cos^{-1}\left[\frac{(7\mi)^{2}+(8\,mi)^{2}-(5\,mi)^{2}}{2\cdot (8\,mi)\cdot (7\,mi)} \right]$

$\theta \approx 38.213^{\circ}$

The bearing needed to navigate from island B to island C is approximately 38.213º.

8 0

3 years ago

Compare the line passing through the points (–2, –9) and (4, 6) to the line given by the equation y = 25x – 4.

elena-s [515]

Answer:

D. They have the same y-intercep

Step-by-step explanation:

Before the comparison will be efficient, let's determine the equation of the two points and the x intercept .

(–2, –9) and (4, 6)

Gradient= (6--9)/(4--2)

Gradient= (6+9)/(4+2)

Gradient= 15/6

Gradient= 5/2

Choosing (–2, –9)

The equation of the line

(Y+9)= 5/2(x+2)

2(y+9)= 5(x+2)

2y +18 = 5x +10

2y =5x -8

Y= 5/2x -4

Choosing (4, 6)

The equation of line

(Y-6)= 5/2(x-4)

2(y-6) = 5(x-4)

2y -12 = 5x -20

2y = 5x-8

Y= 5/2x -4

From the above solution it's clear that the only thing the both equation have in common to the given equation is -4 which is the y intercept

4 0

3 years ago

Which of the following is equivalent to log34

olasank [31]

hope it helps u...

plz mark as brainliest...

4 0

4 years ago

Can someone help me

erik [133]

Answer:

6 < x < 23.206

Step-by-step explanation:

To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.

The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...

(2x -12)/2 = arcsin(16/22×sin(24°))

x ≈ 23.206°

Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...

2x -12 = 0

x -6 = 0 . . . . . divide by 2

x = 6 . . . . . . . add 6

Then the range of values of x will be ...

6 < x < 23.206

_____

* One way to do this is to make use of the law of cosines:

22² = AB² + AC² -2·AB·AC·cos(48°)

16² = AD² + AC² -2·AD·AC·cos(2x-12)

The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...

22² = 2·AB²·(1-cos(48°))

16² = 2·AB²·(1 -cos(2x-12))

The value of AB drops out of the ratio of these equations, and the result for x is as above.

4 0

3 years ago

Read 2 more answers