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3 years ago

What is the slope of this line? A. 3 B. -1/3 C. -3 D. -2 E. 4

Mathematics

2 answers:

MA_775_DIABLO [31]3 years ago

6 0

Answer:

C. -3

Explanation:

Given problem;

To solve for the slope of the line;

The slope of a line is the vertical distance to the horizontal displacement on a line.

Slope = $\frac{Vertical distance}{horizontal displacement}$

For lines such as this, we use the formula below;

Slope = $\frac{y_{2} - y_{1} }{x_{2} - x_{1} }$

To find the slope reference the image attached;

point 1 (0,4) x₁ = 0 y₁ = 4

point 2 (2,-2) x₂ = 2 y₂ = -2

Now input the parameters and solve;

Slope = $\frac{-2 - 4}{2 - (0)}$ = $\frac{-6}{2}$

Slope = -3

The slope of the line is -3

Bingel [31]3 years ago

5 0

Answer:

-3

Step-by-step explanation:

Pick two points on the graph, lets go with (0, 4) and (1, 1)

y = mx + b

m is the slope and to figure it out use below formula

m = (y₂ - y₁) / (x₂ - x₁)

= (1 - 4) / (1 - 0)

= -3 / 1

= -3

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Answer d)

$a= 10*\sqrt{3}$ $b=5*\sqrt{3}$ , $c=15$ , and $d=5$

Step-by-step explanation:

Notice that there are basically two right angle triangles to examine: a smaller one in size on the right and a larger one on the left, and both share side "b".

So we proceed to find the value of "b" by noticing that it the side "opposite side to angle 60 degrees" in the triangle of the right (the one with hypotenuse = 10). So we can use the sine function to find its value:

$b=10*sin(60^o)= 10*\frac{\sqrt{3} }{2} = 5*\sqrt{3}$

where we use the fact that the sine of 60 degrees can be written as: $\frac{\sqrt{3} }{2}$

We can also find the value of "d" in that same small triangle, using the cosine function of 60 degrees:

$d=10*cos(60^o)=10* \frac{1}{2} = 5$

In order to find the value of side "a", we use the right angle triangle on the left, noticing that "a" s the hypotenuse of that triangle, and our (now known) side "b" is the opposite to the 30 degree angle. We use here the definition of sine of an angle as the quotient between the opposite side and the hypotenuse:

$sin(30^o)= \frac{b}{a} \\a=\frac{b}{sin(30)} \\a=\frac{5*\sqrt{3} }{\frac{1}{2} } \\a= 10*\sqrt{3}$

where we used the value of the sine function of 30 degrees as one half: $\frac{1}{2}$

Finally, we can find the value of the fourth unknown: "c", by using the cos of 30 degrees and the now known value of the hypotenuse in that left triangle:

$c=10*\sqrt{3} * cos(30^o)=10*\sqrt{3} *\frac{\sqrt{3} }{2} \\c= 5*3=15$

Therefore, our answer agrees with the values shown in option d)

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