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3 years ago

Find the equation using the y=mx+c method

Mathematics

1 answer:

lesya [120]3 years ago

3 0

Answer:

Step-by-step explanation:

so we know that y=mx+c

c is the point where the line crosses the y-axis

y=mx+0

m is the gradient

the gradient is rise/run

pick any two points

so it is 2/1

2/1 is 2

therefore it is

y=2x+0

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Can anyone help me solve this please?

shutvik [7]

Answer:

The linear equations for the given Rectangle are

$y = 4x +4$
$y = 6x - 6$

Step-by-step explanation:

Given:

area given by 'y'

For figure 1,

$Length = x +1\\Breadth = 4\\\textrm{Area of Rectangle} = Length\times Breadth\\y= (x+1)\times 4\\y= 4x + 4$

For figure 2,

$Length = 2x -2\\Breadth = 3\\\textrm{Area of Rectangle} = Length\times Breadth\\y= (2x-2)\times 3\\y= 6x - 6$

Now we will plot the these two line on DESMOS so, we will have the intersection of these two line.

So, the intersection points are (5,24) that is x coordinate '5' and y coordinate '24'.

∴ $x=5\ and\ y=24$

For figure 1,

∴ $Length =x+1\\=5+1\\=6\ unit\\Area=4x + 4\\=4\times5 +4\\=24\textrm{ square units}$

For figure 2,

∴ $Length =2x-2\\=2\times5-2\\=8\ unit\\Area=6x - 6\\=6\times5 -6\\=24\textrm{ square units}$

7 0

4 years ago

Here is the question 13. Higher Order Thinking In the equation 8x - the variable x represents the same value. Which value of x i

pav-90 [236]

Answer:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ 8x & {0} & {8} & {16} & {24} \ \end{array}$

Step-by-step explanation:

Given

$8x$

Required:

Solve for $x = 0,1,2,3$

For x = 0

$8x = 8 * 0$

$8x = 0$

For x = 1

$8x = 8 *1$

$8x = 8$

For x= 2

$8x = 8 * 2$

$8x = 16$

For x = 3

$8x = 8 *3$

$8x = 24$

Hence, the results are:

$\begin{array}{ccccc}x & {0} & {1} & {2} & {3} \ \\ 8x & {0} & {8} & {16} & {24} \ \end{array}$

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3 years ago

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djverab [1.8K]

Answer:

63.69

Step-by-step explanation:

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77julia77 [94]

Answer:

Step-by-step explanation:

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Christian evaluated the expression, but his teacher told him he made a mistake.

NeX [460]

(24-3) + 18
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3 years ago