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3 years ago

10

What line is perpendicular to the line 6x+y=-4

1 answer:

Arte-miy333 [17]3 years ago

8 0

Answer:

DONT KNOW TO BE HONEST SEVRS ARE SWAMPED

Step-by-step explanation:

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Please help asap!!!

erik [133]

D-A-C-B should be the order
i really hope this helps

6 0

3 years ago

Read 2 more answers

What other points are on the line of direct variation through (5, 12)? Check all that apply. (0, 0) (2.5, 6) (3, 10) (7.5, 18) (

aalyn [17]

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form $y/x=k$ or $y=kx$

in this problem we have

the point $(5,12)$ is on the line of direct variation

so

Find the constant of proportionality k

$y/x=k$ -------> substitute ------> $k=12/5$

the equation is

$y=\frac{12}{5}x$

Remember that

If a point is on the line of direct variation

then

the point must satisfy the equation of direct variation

we're proceeding to verify each point

<u>case A)</u> point $(0,0)$

$x=0\ y=0$

Substitute the value of x and y in the direct variation equation

$0=\frac{12}{5}*0$

$0=0$ -------> is true

therefore

the point $(0,0)$ is on the line of direct variation

<u>case B)</u> point $(2.5,6)$

$x=2.5\ y=6$

Substitute the value of x and y in the direct variation equation

$6=\frac{12}{5}*2.5$

$6=6$ -------> is true

therefore

the point $(2.5,6)$ is on the line of direct variation

<u>case C)</u> point $(3,10)$

$x=3\ y=10$

Substitute the value of x and y in the direct variation equation

$10=\frac{12}{5}*3$

$10=7.2$ -------> is not true

therefore

the point $(3,10)$ is not on the line of direct variation

<u>case D)</u> point $(7.5,18)$

$x=7.5\ y=18$

Substitute the value of x and y in the direct variation equation

$18=\frac{12}{5}*7.5$

$18=18$ -------> is true

therefore

the point $(7.5,18)$ is on the line of direct variation

<u>case E)</u> point $(12.5,24)$

$x=12.5\ y=24$

Substitute the value of x and y in the direct variation equation

$24=\frac{12}{5}*12.5$

$18=30$ -------> is not true

therefore

the point $(12.5,24)$ is not on the line of direct variation

<u>case F)</u> point $(15,36)$

$x=15\ y=36$

Substitute the value of x and y in the direct variation equation

$36=\frac{12}{5}*15$

$36=36$ -------> is true

therefore

the point $(15,36)$ is on the line of direct variation

therefore

<u>the answer is</u>

$(0,0)$

$(2.5,6)$

$(7.5,18)$

$(15,36)$

5 0

3 years ago

Read 2 more answers

assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

6 0

2 years ago

Pls help i dont know ?

Nimfa-mama [501]

Answer:

$8 \times(2+15)$

Step-by-step explanation:

5 0

1 year ago

Read 2 more answers

An exponential function, f, passes through the points (-2,-4) and (1,10). Which two points would lie on the graph of function g

Lubov Fominskaja [6]

Answer:

(-2,-12) and (1,30)

Step-by-step explanation:

we know that

f(x) passes through the points (-2,-4) and (1,10)

That means

For x=-2 ----> f(-2)=-4

For x=1 ----> f(1)=10

we have

g(x)=3f(x)

so

substitute the given value of f(x) to obtain the value of the function g(x)

For x=-2 -----> g(-2)=3f(-2) ----> g(-2)=3(-4)=-12

For x=1 -----> g(1)=3f(1) ----> g(1)=3(10)=30

therefore

The two points that would lie on the graph of function g are

(-2,-12) and (1,30)

3 0

3 years ago