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3 years ago

Find m∠4 if m∠1 = 100°

Mathematics

1 answer:

11111nata11111 [884]3 years ago

7 0

Answer:

obtuse angle

Step-by-step explanation:

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write the x and y coordinate (in the second and third column, respectively) of dilation of quadrilateral ABCD with vertices A(1,

gayaneshka [121]

Answer:

The coordinates of the dillated vertices are $A'(x,y) = (2,2)$ , $B'(x,y) = (4,4)$ , $C'(x,y) = (8,2)$ and $D'(x,y) = (4,-2)$ .

Step-by-step explanation:

From Linear Algebra, we define dilation by the following equation:

$P'(x,y) = O(x,y) + k\cdot [P(x,y)-O(x,y)]$ (1)

Where:

$O(x,y)$ - Center of dilation, dimensionless.

$P(x,y)$ - Original point, dimensionless.

$k$ - Scale factor, dimensionless.

$P'(x,y)$ - Dilated point, dimensionless.

If we know that $O(x,y) = (0, 0)$ , $k = 2$ , $A(x,y) = (1,1)$ , $B(x,y) = (2,2)$ , $C(x,y) = (4,1)$ and $D(x,y) = (2,-1)$ , then the dilated points are, respectively:

Point A

$A'(x,y) = O(x,y) + k\cdot [A(x,y)-O(x,y)]$ (2)

$A'(x,y) = (0,0) + 2\cdot [(1,1)-(0,0)]$

$A'(x,y) = (2,2)$

Point B

$B'(x,y) = O(x,y) + k\cdot [B(x,y)-O(x,y)]$ (3)

$B'(x,y) = (0,0) + 2\cdot [(2,2)-(0,0)]$

$B'(x,y) = (4,4)$

Point C

$C'(x,y) = O(x,y) + k\cdot [C(x,y)-O(x,y)]$

$C'(x,y) = (0,0) + 2\cdot [(4,1)-(0,0)]$

$C'(x,y) = (8,2)$

Point D

$D'(x,y) = O(x,y) + k\cdot [D(x,y)-O(x,y)]$

$D'(x,y) = (0,0) + 2\cdot [(2,-1)-(0,0)]$

$D'(x,y) = (4,-2)$

The coordinates of the dillated vertices are $A'(x,y) = (2,2)$ , $B'(x,y) = (4,4)$ , $C'(x,y) = (8,2)$ and $D'(x,y) = (4,-2)$ .

4 0

3 years ago

How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)

shepuryov [24]

$LHS\\ \\ =\frac { \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } }{ \tan { x } } \\ \\ =\frac { 1 }{ \tan { x } } \left( \tan ^{ 2 }{ x-\sin ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } -\frac { \sin ^{ 2 }{ x\cos ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \right) \\ \\ =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^{ 2 }{ x-\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } }{ \cos ^{ 2 }{ x } } \right)$

$\\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\left( 1-\cos ^{ 2 }{ x } \right) } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^{ 2 }{ x\cdot \sin ^{ 2 }{ x } } }{ \cos ^{ 2 }{ x } } \\ \\ =\frac { \cos { x } \sin ^{ 4 }{ x } }{ \sin { x\cos ^{ 2 }{ x } } } \\ \\ =\frac { \sin ^{ 3 }{ x } }{ \cos { x } }$

$\\ \\ =\sin ^{ 2 }{ x } \cdot \frac { \sin { x } }{ \cos { x } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \frac { \cos { x } }{ \sin { x } } } \\ \\ =\sin ^{ 2 }{ x } \cdot \frac { 1 }{ \cot { x } } \\ \\ =\frac { \sin ^{ 2 }{ x } }{ \cot { x } } \\ \\ =RHS$

8 0

3 years ago

Find the missing length of the triangle.<br> 26 in.<br> a<br> 24 in.

Elis [28]

Answer:

Step-by-step explanation:

26^2-24^2=100

sqrt of 100 is 10

3 0

3 years ago

Read 2 more answers

A cone had a diameter of 6 centimeters and a height that is 5 times the diameter. Using 3.14 for pi, which of the following can