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2 years ago

6

Geometry question: What are the coordinates of the image of the point (3, -11) under a dilation with a scale factor of 5 and the

center of dilation at the origin?

1 answer:

Alexxandr [17]2 years ago

7 0

Answer:

The coordinates are (15,-55).

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ASHA 777 [7]

Answer:

100 revolitions

Step-by-step explanation:

180s = 3min

50s = 1min

50 * 2 = 100

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QveST [7]

$\bf 27\implies 3^3\\\\
i^3\implies i\cdot i\cdot i\implies i^2\cdot i\implies -1\cdot i\implies -i\\\\
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x^3-27i=0\implies x^3+3^3(-i)=0\implies x^3+(3^3i^3)=0
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x^3+(3i)^3=0\\\\
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\textit{difference of cubes}
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(a+b)(a^2-ab+b^2)= a^3+b^3\\\\
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$\bf (x+3i)[x^2-3ix+(3i)^2]=0\implies 
\begin{cases}
x+3i=0\implies \boxed{x=-3i}\\\\
x^2-3ix+(3i)^2=0
\end{cases}
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\textit{now, for the second one, we'd need the quadratic formula}
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3 years ago

. Use the grid on the right.

timofeeve [1]

Answer with Step-by-step explanation:

We are given that

O(0,0),P(3,-1) and Q(2,3) on the coordinate plane.

Mark point O(0,0), P(3,-1) and Q(2,3) on the graph and then join OP,PQ and OQ.

b.We have to find OP and OQ are perpendicular or not.

Slope-formula: $m=\frac{y_2-y_1}{x_2-x_1}$

By using the formula

Slope of OP= $\frac{-1-0}{3-0}=-\frac{1}{3}$

Slope of OQ= $\frac{3-0}{2-0}=\frac{3}{2}$

When two lines are perpendicular then

Slope of one line= $-\frac{1}{slope\;of\;other\;line}$

Slope of line OP $\neq -\frac{1}{slope\;of\;line\;OQ}$

Hence, line OP and OQ are not perpendicular.

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3 years ago

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Aloiza [94]

Answer:

1/1296

Step-by-step explanation:

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Tamiku [17]

Answer:

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