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mestny [16]
2 years ago
14

Graph the equation using the slope and the y intercept 2x-3y=8

Mathematics
1 answer:
sammy [17]2 years ago
7 0

Step-by-step explanation:

<h2>Find slope-intercept structure </h2>
  • y=
  • \frac{2}{3} x -  \frac{8}{3}
  • m=2/3, b=-8/3
  • therefore the slope will be 2/3

To find y-intercept

  • let x=0
  • 2(0)-3y=8
  • -3y=8
  • y=-3/8

To find x-intercept

  • let y=0
  • 2x-3(0)=8
  • 2x=8
  • x=4

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A punch bowl had 2\dfrac122 2 1 ​ 2, start fraction, 1, divided by, 2, end fraction liters of juice in it. Liam drank \dfrac13 3
8_murik_8 [283]

Answer:

5/6 liters.

Step-by-step explanation:

Given that:

Amount of juice in a punch bowl = 2 * 1/2 liters

= 5/2 liters.

Part of juice drank by Liam from the bowl = 1/3

Amount of juice drank by Liam =

= 5/2 * 1/3 = 5\6 liters.

4 0
3 years ago
A conical-shaped umbrella has a radius of 0.4 m and a height of 0.45 m. Calculate the amount of fabric needed to manufacture thi
bonufazy [111]

Answer:

The correct answer will be "0.756596 m²".

Step-by-step explanation:

The given values are:

The radius of an umbrella,

r = 0.4 m

The height of an umbrella,

h = 0.45 m

As we know,

The lateral surface of an umbrella will be:

⇒  \pi r\sqrt{r^2+h^2}

On substituting the values, we get

⇒  \pi \times 0.4\sqrt{(0.4)^2=(0.45)^2}

⇒  0.756596 \ m^2

So that the amount of fabric needed will be "0.756596 m²".

7 0
3 years ago
Find the horizontal and vertical asymptotes of​ f(x). ​f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the
miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* <em>Lets the meaning of vertical and horizontal asymptotes</em>

- <u><em>Vertical asymptotes</em></u> are vertical lines which correspond to the zeroes

  of the denominator of a rational function

- <u><em>A horizontal asymptote</em></u> is a y-value on a graph which a function

 approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

 denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

 denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

 then there is a horizontal asymptote at y = leading coefficient of the

 numerator ÷ leading coefficient of the denominator

* <em>Lets solve the problem</em>

∵ f(x)=\frac{6x}{x+2}

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ <em>The horizontal asymptote can be described by the line y = 6</em>

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ <em>The vertical asymptote can be described by the line x = -2</em>

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Ede4ka [16]

Answer:

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Step-by-step explanation:

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2 years ago
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