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3 years ago

What is the slope of a line perpendicular to line 6x + 3y = 15

Mathematics

1 answer:

NISA [10]3 years ago

8 0

Step-by-step explanation:

Given line is 6x+3y=15

Slope of the given line (m)=-x/y

=-6/3

=-2

let slope of the line perpendicular to the given line be m'.

Condition of perpendicularity,

m×m'=-1

-2×m'=-1

m'=-1/-2

m'=1/2

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PSYCHO15rus [73]

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3 years ago

Miguel has $20 to spend on going to a movie. If he spends $7.25 on a move ticket, $3.95 for snacks, and $1.75 for bus fare, how

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3 years ago

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3 years ago

Read 2 more answers

The diagram shows a 3 cm x 5 cm x 4 cm cuboid

Rainbow [258]

Answer:

a) The length of segment AC is approximately 5.83 centimeters.

b) The angle ACD is approximately 34.5º.

Step-by-step explanation:

a) Since $AB \perp BC$ , the length of segment $AC$ is determined by Pythagorean Theorem, that is:

$AC = \sqrt{(5\,cm)^{2}+(3\,cm)^{2}}$

$AC \approx 5.831\,cm$

The length of segment AC is approximately 5.831 centimeters.

b) Since $AB \perp BC \perp AD$ , the length of segment $AD$ is determined by this Pythagorean identity:

$AD = \sqrt{(3\,cm)^{2}+(5\,cm)^{2}+(4\,cm)^{2}}$

$AD \approx 7.071\,cm$

The angle ACD is determined by the following trigonometric expression:

$\cos C = \frac{AC}{CD}$

$\cos C = \frac{5.831\,cm}{7.071\,cm}$

$\cos C = 0.825$

$C = \cos^{-1} 0.825$

$C \approx 34.448^{\circ}$

The angle ACD is approximately 34.448º.

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3 years ago