All categories

3 years ago

Which of the following equations is paralllel to the line y+3x+1 and passes through the point (6,-3)

Mathematics

1 answer:

RideAnS [48]3 years ago

7 0

Answer:

y=3x-21

Step-by-step explanation:

a line is parallel if the slopes of the lines are the same

so that means that out line should look something like this

y=3x+b

then, we use our point given to find b

-3=18+b

b=-21

the equation of out line is y=3x-21

You might be interested in

Если x=6 то y=<br>если y=-4 то x=

jok3333 [9.3K]

Step-by-step explanation:

ssshzvecucsussddyz wksoav

8 0

3 years ago

Read 2 more answers

15 point help asap marking brainlest!!!!!!!!!!!!!!

ziro4ka [17]

slope =(y2-y1)/(x2-x1)

m=(-1-5)/(1--1)

m=-6/2

m=-3

y-y1=m(x-x1) point slope form

y-5 = -3 (x--1)

y-5 = -3 (x+1)

distribute

y-5 = -3x -3

add 5 to each side

y = -3x+2

Choice D

4 0

3 years ago

Read 2 more answers

You have four ice cubes in your cup of lemonade each ice cube has a length of 2 1/4 cm a width of 2 2/3 cm and a height of 2 cm

Margarita [4]

Answer: 6

Step-by-step explanation:

4 0

3 years ago

when 6 is subtracted from the square of a number, the result is 5 times the number. Find the negative solution.

sveta [45]

When 6 is subtracted from the square of a number, the result is 5 times the number, then the negative solution is -1

<h3><u>Solution:</u></h3>

Given that when 6 is subtracted from the square of a number, the result is 5 times the number

To find: negative solution

Let "a" be the unknown number

Let us analyse the given sentence

square of a number = $a^2$

6 is subtracted from the square of a number = $a^2 - 6$

5 times the number = $5 \times a$

<em><u>So we can frame a equation as:</u></em>

6 is subtracted from the square of a number = 5 times the number

$a^2 - 6 = 5 \times a\\\\a^2 -6 -5a = 0\\\\a^2 -5a -6 = 0$

<em><u>Let us solve the above quadratic equation</u></em>

For a quadratic equation $ax^2 + bx + c = 0$ where $a \neq 0$

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here in this problem,

$a^2-5 a-6=0 \text { we have } a=1 \text { and } b=-5 \text { and } c=-6$

Substituting the values in above quadratic formula, we get

$\begin{array}{l}{a=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2 \times 1}} \\\\ {a=\frac{5 \pm \sqrt{25+16}}{2}=\frac{5 \pm \sqrt{49}}{2}} \\\\ {a=\frac{5 \pm 7}{2}}\end{array}$

We have two solutions for "a"

$\begin{array}{l}{a=\frac{5+7}{2} \text { and } a=\frac{5-7}{2}} \\\\ {a=\frac{12}{2} \text { and } a=\frac{-2}{2}}\end{array}$

We have asked negative solution. So a = -1

Thus the negative solution is -1

6 0

3 years ago

- The four-digit number 1210 has an interesting property.

vampirchik [111]

Answer:

ABCDEFG = 3211000

Step-by-step explanation:

A counts the number of zeroes, and there are 3 zeroes (E, F, G), so A = 3.B counts the number of ones, and there are 2 ones (C, D), so B = 2.C counts the number of twos, and there is 1 two (B), so C = 1.D counts the number of threes, and there is 1 three (A), so D = 1.There is no four, five, six in, so E, F, G are all zeroes.

5 0

3 years ago