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2 years ago

Determine the standard form of the equation of the line that passes through (6, -9) and (-8, 6)

Mathematics

1 answer:

horrorfan [7]2 years ago

4 0

9514 1404 393

Answer:

A. 15x +14y = -36

Step-by-step explanation:

Since we are given two points, we can start with the 2-point form of the equation for a line.

y = (y2 -y1)/(x2 -x1)(x -x1) +y1

y = (6 -(-9))/(-8 -6)(x -6) +(-9)

y = 15/-14(x -6) -9

Multiplying by -14, we have ...

-14y = 15x -90 +126

Adding 14y-36 to both sides gives ...

-36 = 15x +14y . . . . matches choice A

The standard-form equation is ...

15x +14y = -36

_____

Additional comments

It can be easier to start with the form ...

(Δy)x -(Δx)y = (Δy)x1 -(Δx)y1 . . . . . where Δx = x2-x1 and Δy = y2-y1

This gives ...

(6+9)x -(-8-6)y = 15(6) +14(-9)

15x +14y = -36 . . . simplified

You can also start with the slope-intercept form or the point-slope form, if you're more familiar with those. The result will be the same. I find it handy to be familiar with a number of different forms of the equation for a line.

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Describe and correct the error below.

statuscvo [17]

Step-by-step explanation:

It is given that,

-2.2 × 3.7 = 8.14

The first number is -2.2

The second number is 3.7

We know that, (-) × (+) = - (the answer is a negative)

If we multiply -2.2 and 3.7, we will get -8.14.

But in the given case, the answer is 8.18 instead of -8.14. Hence, this is the only error.

8 0

2 years ago

How many times longer is the wavelength of a sound wave with a frequency of 20 waves per second than the wavelength of a sound w

sergejj [24]

The sound wave with a frequency of 20 waves/sec is 800 longer than the wavelength of a sound wave with a frequency of 16,000 waves/sec

<h3>Calculating wavelength </h3>

From the question, we are to determine how many times longer is the first sound wave compared to the second sound water

Using the formula,

v = fλ

∴ λ = v/f

Where v is the velocity

f is the frequency

and λ is the wavelength

For the first wave

f = 20 waves/sec

Then,

λ₁ = v/20

For the second wave

f = 16,000 waves/sec

λ₂ = v/16000

Then,

The factor by which the first sound wave is longer than the second sound wave is

λ₁/ λ₂ = (v/20) ÷( v/16000)

= (v/20) × 16000/v)

= 16000/20

= 800

Hence, the sound wave with a frequency of 20 waves/sec is 800 longer than the wavelength of a sound wave with a frequency of 16,000 waves/sec

Learn more on Calculating wavelength here: brainly.com/question/16396485

#SPJ1

6 0

1 year ago

KNOWLEDGE CHECK: ALEKS

Dmitrij [34]

a and b: decreasing

b and c: constant

c and d: decreasing

d and e: increasing

5 0

2 years ago

Please help me guys as fast as possible I BEG YOU, thanks!

LuckyWell [14K]

Answer:

2u + v - 4w = <40 , -4>

6u - 8v = <-78 , 58>

4v - 7w = <101 , -36>

11u + 3w = <-88 , 89>

Step-by-step explanation:

* Lets find the value of each operation to find its resultant vector

# 2u + v - 4w

∵ u = <-5 , 7>

∴ 2u = <-10 , 14>

∵ v = <6 , -2>

∵ w = <-11 , 4>

∴ 4w = <-44 , 16>

∴ 2u + v - 4w = <-10 + 6 - -44 , 14 + -2 - 16>

∴ 2u + v - 4w = <40 , -4>

# 6u - 8v

∵ u = <-5 , 7>

∴ 6u = <-30 , 42>

∵ v = <6 , -2>

∵8v = <48 , -16>

∴ 6u - 8v = <-30 - 48 , 42 - -16>

∴ 6u - 8v = <-78 , 58>

# 4v - 7w

∵ v = <6 , -2>

∴ 4v = <24 , -8>

∵ w = <-11 , 4>

∴ 7w = <-77 , 28>

∴ 4v - 7w = <24 - -77 , -8 - 28>

∴ 4v - 7w = <101 , -36>

# 11u + 3w

∵ u = <-5 , 7>

∴ 11u = <-55 , 77>

∵ w = <-11 , 4>

∴ 3w = <-33 , 12>

∴ 11u + 3w = <-55 + -33 , 77 + 12>

∴ 11u + 3w = <-88 , 89>

7 0

2 years ago

Pleaseee help meeeee question 1 and 2!!!!!

madam [21]

Answer:

I know #1 is a

Step-by-step explanation:

#1 is a because the bus stops twice reaching zero speed to pick up riders then he stops again but since he has to slow down it's not a instant stop so it gradually decelerates twice but since it has a speed it has a constant line (straight) another easy way to tell us to look at the speed and then look the the graph and see if they match so you can elimnate a answer

3 0

2 years ago