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Gennadij [26K]
2 years ago
13

PLZ PLZ HELP ME ILL MARK BRAINLIEST AND GET A FOLLOW BUT PLZ HELP ME ​

Mathematics
2 answers:
Nesterboy [21]2 years ago
6 0

Answer:

  • See below

Step-by-step explanation:

<u>Initial coordinates and transformations:</u>

  • (x, y) → (4x, 4y) → (3x, 3y)
  • A(-1, -1) → A'(-4, -4) → A''(-3, -3)
  • B(1, 1) → B'(4, 4) → B''(3, 3)
  • C(2, 0) → C'(8, 0) → C''(6, 0)
Korvikt [17]2 years ago
6 0
What the guy said is 100% right
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3 years ago
Need help with my homework ​
Volgvan

Answer:

\displaystyle y=\frac{16-9x^3}{2x^3 - 3}

\displaystyle y=-\frac{56}{13}

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3

Simplify the radical with the cube:

\displaystyle x^3=\frac{3y+16}{2y+9}

Multiply by 2y+9

\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)

Simplify:

\displaystyle x^3(2y+9)=3y+16

Operate the parentheses:

\displaystyle x^3(2y)+x^3(9)=3y+16

\displaystyle 2x^3y+9x^3=3y+16

Subtract 3y and 9x^3:

\displaystyle 2x^3y - 3y=16-9x^3

Factor y out of the left side:

\displaystyle y(2x^3 - 3)=16-9x^3

Divide by 2x^3 - 3:

\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}

ii) To find y when x=2, substitute:

\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}

\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}

\displaystyle y=\frac{16-72}{16- 3}

\displaystyle y=\frac{-56}{13}

\mathbf{\displaystyle y=-\frac{56}{13}}

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2 years ago
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