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zlopas [31]
3 years ago
8

Help with question 24. Please provide explanation! Big points

Mathematics
2 answers:
DerKrebs [107]3 years ago
6 0

Answer:

4/3

Step-by-step explanation:

-3y + 4x = -6

-3y = -6 -4y

slove for Y

y = -6/-3 -4/-3 x

Y = -2 +4/3x

Y = 4/3 x -2

Y = 4/3 x -2

so slope or gradient is coefficient of x

4/3 is gradient.

DochEvi [55]3 years ago
3 0

Answer:

a

Equation of a straight line= mx+c

where m is the gradient...

-3y + 4x = -6

  • Rearrange this equation to the form mx+c

-3y = -4x - 6

y = -4/-3x - 6/3

Simplifying we get y = 4/3x - 2

Therefore 4/3 is the gradient because according to the equation y = mx+c, your m value (gradient) is 4/3.

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Determine the midpoint between the two points x(4,-6) and y(-2,8)
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Answer:

p(a, b) = (1, 1)

Step-by-step explanation:

Midpoint formula is

p(a, b)=(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} ) ---------------(1)

Here

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HELPPP PLSSS!!!!!!!!!!! Find the volume of the cone.
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Answer:

The volume of cone is \boxed{\tt{167.47}} unit³.

Step-by-step explanation:

<u>Solution</u> :

As per given question we have provided :

  • ➝ Radius of cone = 4 units
  • ➝ Height of cone = 10 units

Here's the required formula to find the volume of cone :

{\longrightarrow{\pmb{\sf{V_{(Cone)} = \dfrac{1}{3}\pi{r}^{2}h}}}}

  • V = Volume
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  • r = radius
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Substituting all the given values in the formula to find the volume of cone :

{\longrightarrow{\sf{Volume_{(Cone)} = \dfrac{1}{3}\pi{r}^{2}h}}}

{\longrightarrow{\sf{Volume_{(Cone)} = \dfrac{1}{3} \times 3.14{(4)}^{2}10}}}

{\longrightarrow{\sf{Volume_{(Cone)} = \dfrac{1}{3} \times 3.14{(4 \times 4)}10}}}

{\longrightarrow{\sf{Volume_{(Cone)} = \dfrac{1}{3} \times 3.14{(16)}10}}}

{\longrightarrow{\sf{Volume_{(Cone)} = \dfrac{1}{3} \times 3.14 \times 16 \times 10}}}

{\longrightarrow{\sf{Volume_{(Cone)} = \dfrac{1}{3} \times 3.14 \times 160}}}

{\longrightarrow{\sf{Volume_{(Cone)} = \dfrac{1\times 3.14 \times 160}{3}}}}

{\longrightarrow{\sf{Volume_{(Cone)} = \dfrac{3.14 \times 160}{3}}}}

{\longrightarrow{\sf{Volume_{(Cone)} = \dfrac{502.4}{3}}}}

{\longrightarrow{\sf{Volume_{(Cone)}  \approx 167.47}}}

\star{\underline{\boxed{\sf{\purple{Volume_{(Cone)} \approx 167.47\:  {unit}^{3}}}}}}

Hence, the volume of cone is 167.47 unit³.

\rule{300}{2.5}

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