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3 years ago

What is the equation of the line that passes through 1,4 and 2,1

Mathematics

1 answer:

sashaice [31]3 years ago

3 0

Answer:

y = -3x + 7

Step-by-step explanation:

The equation of a line

y = mx + c

y - intercept point y

m - slope of the line

x - intercept point x

c - intercept point of the line

Step 1: find the slope

m = y2 - y1 / x2 - x1

Given two points

( 1 , 4) ( 2 , 1)

x1 = 1

y1 = 4

x2 = 2

y2 = 1

insert the values

m = 1 - 4 / 2 - 1

m = -3/1

m = -3

y = -3x + c

Step 2: substitute any of the two points given into the equation of a line

y = -3x + c

( 1 ,4)

x = 1

y = 4

4 = -3(1) + c

4 = -3 + c

4 + 3 = c

c = 7

Step 3: sub c into the equation

y = -3x + 7

The equation of the line is

y = -3x + 7

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Which phrase describes the linear relationship between the x and y values shown in the table?

kompoz [17]

9514 1404 393

Answer:

C. y is 6 less than x

Step-by-step explanation:

It is not hard to check.

A. 6 times x is 6×8 = 48, not 2

B. 6 times y is 6×2 = 12, not 8

C. 6 less than 8 is 2; 6 less than 9 is 3

D. 6 more than 8 is 14, not 2

The relation described in C matches the table.

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2 years ago

Evaluate the expression for the given value of variable(s)

Olenka [21]

You would:
(4 * 2 - 4) + (3 - 2^2) + (2 * 2^3)
(8 - 4) + (3 - 4) + (16)
4 + -1 + 16 = 19

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3 years ago

Applying properties of Exponents in exercise,use the properties of exponents to simplify the expression.See example 1.

evablogger [386]

Answer: a) 15625, b) 0.2, c) 625, d) 0.008.

Step-by-step explanation:

Since we have given that

a) $(5^2)(5^3)$

As we know that

$a^m\times a^n=a^{m+n}$

So, it becomes,

$5^2\times 5^3=5^{2+3}=5^6=15625$

(b) $(5^2)(5^{-3})$

So, it becomes,

$5^2\times 5^{-3}=5^{2-3}=5^{-1}=\dfrac{1}{5}=0.2$

As we know that

$(a^m)^n=a^{mn}$

So, it becomes,

$5^{2\times 2}=5^4=625$

(d) $5^{-3}$

$a^{-m}=\dfrac{1}{a^m}\\\\5^{-3}=\dfrac{1}{5^3}=\dfrac{1}{125}=0.008$

Hence, a) 15625, b) 0.2, c) 625, d) 0.008.

4 0

3 years ago

How do I factor out a constant before factoring out a quadratic? <br>Factor completely: 3w^2-3w-90

pantera1 [17]

$3w^2-3w-90 =0 \ \ \:3 \\ \\w^2- w-30 =0 \\ \\a=1 , \ b= -1 , c= -30 \\ \\\Delta =b^2-4ac = (-1)^2 -4\cdot1\cdot (-30) =1+120=121 \\ \\w_{1}=\frac{-b-\sqrt{\Delta} }{2a}=\frac{1-\sqrt{121}}{2 }=\frac{ 1-11}{2}=\frac{-10}{2}=-5$

$w_{2}=\frac{-b+\sqrt{\Delta} }{2a}=\frac{1+\sqrt{121}}{2 }=\frac{ 1+11}{2}=\frac{12}{2}=6 \\ \\ Answer:\\ \\ w^2- w-30 =(x+5)(x-6)$

3 0

3 years ago

Find an equation of the plane with x-intercept a, y-intercept b, and z-intercept c. What is the distance between the origin and

Elenna [48]

Answer:

The equation is:

(1/a)x + (1/b)y + (1/c)z = 1

Step-by-step explanation:

The direction vector between the points (a, 0, 0) and (0, b, 0) is given as:

<0 - a, b - 0, 0 - 0>

<-a, b, 0> .....................(1)

The direction vector between (0, 0, c) and (0, b, 0) is given as:

<0 - 0, b - 0, 0 - c>

= <0, b, -c> .....................(2)

To obtain the direction vector that is normal to the surface of the plane, we take the cross product of (1) and (2).

Doing this, we have:

<-a, b, 0> × <0, b, -c> = <-bc, -ac, -ab>

To find the scalar equation of the plane we can use any of the points that we know. Using (0, b, 0), we have:

(-bc)x + (-ac)y + (-ab)z = (-bc)0 + (-ac)b + (-ab)0

(bc)x + (ac)y + (ab)z = (ac)b

Dividing both sides by abc, we have:

(1/a)x + (1/b)y + (1/c)z = 1

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3 years ago