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2 years ago

Hat is the gradient of the line parallel to the line 4y+6x=8?

Mathematics

2 answers:

Kamila [148]2 years ago

6 0

Answer:

Step-by-step explanation:

y = mx + b

m = slope be = y-intercept.

Parallel lines have the same slope.

since m = 6 then slope of this line = 6 and the parallel line will also = 6

____ [38]2 years ago

6 0

Answer:

gradient = - $\frac{3}{2}$

Step-by-step explanation:

Parallel lines have equal slopes/ gradients

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Given

4y + 6x = 8 ( subtract 6x from both sides )

4y = - 6x + 8 ( divide terms by 4 )

y = - $\frac{6}{4}$ x + $\frac{8}{4}$

y = - $\frac{3}{2}$ x + 2 ← in slope- intercept form

with slope / gradient = - $\frac{3}{2}$

The gradient of a parallel line is then m = - $\frac{3}{2}$

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Step-by-step explanation:

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3 years ago

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Troyanec [42]

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3 years ago

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14 out of the 25 students passed the quiz. What percent of students passed the quiz ?

Mnenie [13.5K]

Answer:

Step-by-step explanation:

A fraction consists of two numbers and a fraction bar:

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Hoped it helped

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3 years ago

Solve this for me please I will mark brainliest?

Alexandra [31]

Answer:

There were 53 cakes ordered.

Step-by-step explanation:

☆Cross multiply.

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▪Happy to help <3

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3 years ago

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NEED ASAP plz. I will also mark brainiest! Don't just take points, I will report you.

PilotLPTM [1.2K]

Part A:

You may choose the two lines connecting the origin and points A and B, and choose the portion of the space between them.

The line between the origin and A is

$y = 3x$

We want everything below this line (line included), so the first inequality is

$y \leq 3x$

The line between the origin and B is

$y = \dfrac{1}{3}x$

We want everything above this line (line included), so the second inequality is

$y \geq \dfrac{1}{3}x$

Create a system with these two inequalities and you'll have an area including only points A and B

Part B:

To verify the solutions, we can plug the coordinates of A and B in this system and check that we get something true: the coordinates of point A are (1,3), while the coordinates of point B are (3,1). The system becomes:

$A:\begin{cases}3 \leq 3\cdot 1\\3 \geq \frac{1}{3}\cdot 1\end{cases},\quad B:\begin{cases}1 \leq 3\cdot 3\\1 \geq \frac{1}{3}\cdot 3\end{cases}$

Which means

$A:\begin{cases}3 \leq 3\\3 \geq \frac{1}{3}\end{cases},\quad B:\begin{cases}1 \leq 9\\1 \geq 1\end{cases}$

And these are all true. So, the system is satisfied, which means that the points belong to the shaded area.

Part C

If you draw the line, you'll see that the only points that lay below the line are B and C. In fact, if we plug the coordinates we have

$B:\ 1$

And this are both true. You can check the coordinates of all other points, and see that they won't satisfy the inequality y<3x-6

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3 years ago