All categories

3 years ago

Does the equation y = x(x - 3) represent a linear function

Mathematics

1 answer:

Flauer [41]3 years ago

7 0

Answer:

Step-by-step explanation:

No. You can see why if you remove the brackets.

y=x^2 - 3x.

A linear function must have x and y with powers of 1. Also it can't something like x^2 - y^2 = 5

x^2 - y^2 = 0 is the only exception I can think of, but that really is a system of equations. If you graph it. you get two equations which cross at 0 and nowhere else.

You might be interested in

What is the value of x?<br><br> x = ___

Shalnov [3]

Answer:

x = 24

Step-by-step explanation:

The triangle is split into two similar triangles. They have the same shape but not the same size. To solve, write a proportion.

$\frac{56}{44.8}=\frac{35}{x+4}$

Cross multiply to solve for x.

56(x+4) = 44.8(35)

56x + 224 = 1,568

56x = 1,344

x= 24

4 0

3 years ago

A rectangle has one side that measures 3x + 1 and another that measures 4y + 3. Keith says that the perimeter of the rectangle i

mart [117]

Answer:

he added the dimensions of the rectangle. He used the wrong formula. the formula he used was length + breadth. He should have multiplied by 2

6x + 8y +8

Step-by-step explanation:

Perimeter of a rectangle = 2 x (L + B)

L = length

b = breadth

2 (3x + 4y + 4. ) = 6x + 8y +8

5 0

3 years ago

Can someone help me plz

prisoha [69]

Answer:

-3/-6, 2/2

Step-by-step explanation:

I'm not really sure how else to finish this

3 0

3 years ago

What is the slope of 6x-3y=18??

dolphi86 [110]

$the \ slope \ intercept \ form \ is : \\ \\ y= mx +b \\ \\6x-3y=18 \\ \\-3y=-6x+18 \ \ /:(-3)\\ \\y=\frac{-6}{-3}x + \frac{18}{-3} \\ \\y=2x-6 \\ \\ m = 2$

6 0

3 years ago

Find the directional derivative of f(x,y,z)=2z2x+y3f(x,y,z)=2z2x+y3 at the point (−1,4,3)(−1,4,3) in the direction of the vector

Feliz [49]

$f(x,y,z)=2z^2x+y^3$

$f$ has gradient

$\nabla f(x,y,z)=2z^2\,\vec\imath+3y^2\,\vec\jmath+4xz\,\vec k$

which at the point (-1, 4, 3) has a value of

$\nabla f(-1,4,3)=18\,\vec\imath+48\,\vec\jmath-12\,\vec k$

I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say $\vec u=15\,\vec\imath+25\,\vec\jmath$ , in which case we have

$\|\vec u\|=\sqrt{15^2+25^2}=5\sqrt{34}$

Then the derivative of $f$ at (-1, 4, 3) in the direction of $\vec u$ is

$D_{\vec u}f(-1,4,3)=\nabla f(-1,4,3)\cdot\dfrac{\vec u}{\|\vec u\|}=\boxed{\dfrac{294}{\sqrt{34}}}$

4 0

3 years ago