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2 years ago

Write the equation of a line that is

Mathematics

1 answer:

il63 [147K]2 years ago

8 0

Answer:

Y= 3x-7

Step-by-step explanation:

Since the line is parallel , the slope is same i.e. 3 .

now the equation should be y= Mx + b where m is the slope and c is the y-intercept

plugging the value of m=3 and ( 3,2) into the equation

2= 3(3) + c

2= 9+ c

c= -7

hence the equation of the line is y=3x-7

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Jason states that Triangle A B C is congruent to triangle R S T. Kelley states that Triangle A B C is congruent to triangle T S

Brilliant_brown [7]

Answer:

The correct option is;

Jason's statement is correct. RST is the same orientation, shape, and size as ABC

Step-by-step explanation:

Here we have

ABC = (2, 1), (3, 3), (4, 1)

RST = (-4, -2), (-3, 0), (-2, -2)

Therefore the length of the sides are as follows

AB = $\sqrt{(2-3)^2+(1-3)^2} = \sqrt{5}$

AC = $\sqrt{(2-4)^2+(1-1)^2} =2$

BC = $\sqrt{(3-4)^2+(3-1)^2} = \sqrt{5}$

For triangle SRT we have

RS = $\sqrt{(-4-(-3))^2+(-2-0)^2} = \sqrt{5}$

RT = $\sqrt{(-4-(-2))^2+(-2-(-2))^2} = 2$

ST = $\sqrt{(-3-(-2))^2+(0-(-2))^2} = \sqrt{5}$

Therefore their dimensions are equal

However the side with length 2 occurs between (2, 1) and (4, 1) in triangle ABC and between (-4, -2) and (-2, -2) in triangle RST

That is Jason's statement is correct. RST is the same orientation, shape, and size as ABC.

4 0

3 years ago

Read 2 more answers

Who came up with y=mx+b

allsm [11]

Answer:Renee Descartes was the person that invented slope of a line

Step-by-step explanation:Renee Descartes was the person that invented slope of a line

6 0

3 years ago

What is the solution for -10 < x - 9?

enyata [817]

-10 < x - 9

-10+9 < x - 9+9

-1<x, or
x> - 1

4 0

3 years ago

What is the solution of x-7= ir+14,

weqwewe [10]

Answer:

x=12

Step-by-step explanation:

put the x at one side and the numbers ar the other

5/2x-3/4x=14+7

10/4x-3/4x=21

7/4x=21

7x=84

x=12

3 0

3 years ago

Ex 2.8 3. find the maximum value of y for the curve y=x^5 -3 for -2≤x≤1

harkovskaia [24]

$y=x^5-3\\ y'=5x^4\\\\ 5x^4=0\\ x=0\\ 0\in [-2,1]\\\\ y''=20x^3\\\\
y''(0)=20\cdot0^3=0$

The value of the second derivative for $x=0$ is neither positive nor negative, so you can't tell whether this point is a minimum or a maximum. You need to check the values of the first derivative around the point.
But the value of $5x^4$ is always positive for $x\in\mathbb{R}\setminus \{0\}$ . That means at $x=0$ there's neither minimum nor maximum.
The maximum must be then at either of the endpoints of the interval $[-2,1]$ .
The function $y$ is increasing in its entire domain, so the maximum value is at the right endpoint of the interval.

$y_{max}=y(1)=1^5-3=-2$

4 0

3 years ago