Someone help me with this one

Mathematics

2 answers:

Gala2k [10]3 years ago

6 0

Answer:

Subtract 1/3 from 2/3 dont forget to add the variable h!

1/3h + 11 = 8

-11 -11

1/3h = -3

Now we multiply by 3 on both sides

H = -9

9966 [12]3 years ago

6 0

H = -9 tell me if you need me to do the work

You might be interested in

What is the domain (in interval notation) of the following functions?

alexgriva [62]

$1.\\g(x)=\frac{3}{5x-4}\\\\D:5x-4\neq0\to5x\neq4\ \ \ /:5\to x\neq\frac{4}{5}\to x\in\mathbb{R}\ \backslash\ \{\frac{4}{5}\}\\\\2.\\h(x)=\frac{\sqrt{x}}{x-5}\\\\D:x\geq0\ \wedge\ x-5\neq0\to x\geq0\ \wedge\ x\neq5\to x\in\left$

$3.\\f(x)=\frac{\sqrt{x}}{x^2-5x}\\\\D:x\geq0\ \wedge\ x^2-5x\neq0\to x\geq0\ \wedge\ x(x-5)\neq0\\\\\to x\geq0\ \wedge\ x\neq0\ \wedge\ x\neq5\to x\in\mathbb{R^+}\ \backslash\ \{5\}\\\\4.\\g(x)=\frac{\sqrt{x}+5}{x^2-x-20}\\\\D:x\geq0\ \wedge\ x^2-x-20\neq0\to x\geq0\ \wedge\ (x+4)(x-5)\neq0\\\\\to x\geq0\ \wedge\ x\neq-4\ \wedge\ x\neq5\to x\in\left$

$5.\\h(x)=\frac{3}{x^2+1}\\\\D:x^2+1\neq0\to x^2\neq-1\to x\in\mathbb{R}\\\\6.\\f(x)=\frac{\sqrt{x-2}}{x+1}\\\\D:x-2\geq0\ \wedge\ x+1\neq0\to x\geq2\ \wedge\ x\neq-1\to x\in\left$

$7.\\g(x)=\frac{x^2}{3x^2-x-2}\\\\D:3x^2-x-2\neq0\to (3x+2)(x-1)\neq0\to x\neq-\frac{2}{3}\ \wedge\ x\neq1\\\\\to x\in\mathbb{R}\ \backslash\ \{-\frac{2}{3};\ 1\}\\\\8.\\h(x)=3(x-4)^2-7\\\\D:x\in\mathbb{R}$

4 0

3 years ago

Helpppp pleaseee i need a quick answer

aleksandr82 [10.1K]

Answer:

32 i think

Step-by-step explanation:

its 16 cm, so 2 full rotations would be 16 times 2

which is 32

4 0

3 years ago

Read 2 more answers