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SIZIF [17.4K]
3 years ago
8

Please help me please

Mathematics
1 answer:
Taya2010 [7]3 years ago
3 0
The answer should be lead designer <span />
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How do you dilate a figure on the coordinate plane and determine its scale factor?
levacccp [35]

Answer:

All you have to do is scale up the coordinates of the plane by the factor. Multiply the values of each coodinate by the scale factor.

Step-by-step explanation:

3 0
3 years ago
What is the slope of the line that passes through the points (-4, -1) and (4, 5)?
andreyandreev [35.5K]

Answer:

3/4

Step-by-step explanation:

Since we have two points, we can use the slope formula

m = ( y2-y1)/(x2-x1)

   = ( 5 - -1)/( 4 - -4)

   = ( 5+1)/(4+4)

   = 6/8

   = 3/4

7 0
2 years ago
2 tan 30°<br>II<br>1 + tan- 300​
shusha [124]

Question:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}

Answer:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}= sin(60^{\circ})

Step-by-step explanation:

Given

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}

Required

Simplify

In trigonometry:

tan(30^{\circ}) = \frac{1}{\sqrt{3}}

So, the expression becomes:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}

Simplify the denominator

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}

Express the fraction as:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}= \frac{2}{\sqrt 3} / \frac{4}{3}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{2}{\sqrt 3} * \frac{3}{4}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{1}{\sqrt 3} * \frac{3}{2}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{3}{2\sqrt 3}

Rationalize

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{3\sqrt{3}}{2* 3}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\sqrt{3}}{2}

In trigonometry:

sin(60^{\circ}) =  \frac{\sqrt{3}}{2}

Hence:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}= sin(60^{\circ})

3 0
3 years ago
3/5 of those attending a picnic decided to play touch football. one more decided to play, there were 16 players. how many attend
Effectus [21]
The answer is 9.6
3/5 of 16=9.6
7 0
3 years ago
Rectangle ABCD has vertex coordinates A(1,-2), B(4, -2), C(4,-4), and D(1,
dimulka [17.4K]

Answer:(7,-3)

Step-by-step explanation:

6 0
2 years ago
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