Answer:
take the answer on picture which I attached with your question
Substitute the values of the radius <span>(<span><span>r=3</span>),</span></span><span> the </span>height <span>(<span><span>h=11</span>),</span></span><span> and an approximation of </span>Pi <span>(<span>3.14)</span></span><span> into the </span>formula<span> to find the </span>volume<span> of the </span>cone<span>.
</span><span>V≈+<span>1/3</span>⋅3.14⋅<span>3^2</span>⋅11
</span><span>Simplify.
</span>Write 3.14<span> as a </span>fraction<span> with </span>denominator <span>1.</span><span>
</span><span>V≈+<span>1/3</span>⋅<span>3.14/1</span></span> <span>
</span>Combine <span>1/3</span><span> and </span><span>3.14/1</span><span> to get </span><span><span>3.14/3</span>.</span><span>
</span><span>V≈+<span>3.14/3</span></span><span>
</span>
Replace back in to larger expression<span>.
</span><span>V≈+<span>3.14/3</span>⋅<span>3^2</span>⋅11</span><span>
</span>
Divide 3.14<span> by </span>3<span> to get </span><span>1.04666667.
</span><span>V≈+1.04666667⋅<span>3^2</span>⋅11
</span>Raise 3<span> to the </span>power<span> of </span>2<span> to get </span><span>9.
</span><span>V≈+1.04666667⋅9⋅11
</span>Multiply 1.04666667<span> by </span>9<span> to get </span><span>9.42.
</span><span>V≈+9.42⋅11
</span>Multiply 9.42<span> by </span>11<span> to get </span><span>103.62.
</span><span>V≈+103.62
</span>Remove trailing zeros<span> from </span><span>103.62.
</span><span>V≈+103.62</span><span>f<span>t^3
Volume:
</span></span>V≈+103.62ft^3
Answer:
parallel bc they have same gradient
Step-by-step explanation:
If it is a single transformation, it will need to be a homothetic transformation.
None of the points remain at the old place, so it cannot be a scaling problem with respect to one of the existing points.
A homothetic transformation is bacically a scaling problem, with respect to an arbitrary point called the homothetic centre.
The centre, O, if it exists, is along the point joining any original point and the transformed point.
Here, take a pencil (imaginary one if you wish) and join points AA', BB', CC', DD', EE' and you will find that they are concurrent at point O (3,-6).
So O(3,-6) is the centre of homothety.
The scale factor, as usual, is AO/A'O, or BO/B'O... for transformation from X to Y (X is ABCDE), or the reciprocal if it is from Y to X.
