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3 years ago

Find the slope of a line perpendicular to 2x-y-16

Mathematics

1 answer:

Alekssandra [29.7K]3 years ago

6 0

put the equation 2x - y = 16 in the form of y = mx + c

2x - y = 16

2x = 16 + y

y = 2x - 16

the slope of this line is 2. the slope of a line perpendicular to it would be the negative reciprocal of 2. in other words, it would multiply with 2 to give -1.

you can form this equation with that info

2x = -1

x = -1/2

you can flip and change the sign (numerator) of 2/1

2/1

= -1/2

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Standard Complex Form : $-\frac{25\sqrt{3}}{2}+\frac{25}{2}i$

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We want to rewrite this expression in standard complex form. Let's first evaluate cos(5π / 6). Remember that cos(x) = sin(π / 2 - x). Therefore,

cos(5π / 6) = sin(π / 2 - 5π / 6),

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Now let's evaluate the sin(5π / 6). Similar to cos(x) = sin(π / 2 - x), sin(x) = cos(π / 2 - x). So, sin(5π / 6) = cos(- π / 3). Now let's further simplify from here,

cos(- π / 3) = cos(π / 3)

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Through substitution we receive the expression 25( - √3 / 2 + i(1 / 2) ). Further simplification results in the following expression. <u>As you can see your solution is option a.</u>

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3 years ago

What is the length of segment TR with T(-4,5) and R(2, -1)

Brilliant_brown [7]

Answer:

The length of segment TR with T(-4,5) and R(2, -1) is 8.49

Step-by-step explanation:

We need to find the length of segment TR with T(-4,5) and R(2, -1)

The length of segment can be found using distance formula.

The distance formula is: $Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Putting values and finding length

We have: $x_1=-4, y_1=5, x_2=2, y_2=-1$

$Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance = \sqrt{(2-(-4))^2+(-1-5)^2}\\Distance = \sqrt{(2+4)^2+(-1-5)^2}\\Distance = \sqrt{(6)^2+(-6)^2}\\Distance = \sqrt{36+36}\\Distance = \sqrt{72}\\Distance=8.49$

We found Distance = 8.49

So, The length of segment TR with T(-4,5) and R(2, -1) is 8.49.

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3 years ago