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adelina 88 [10]
3 years ago
9

2 quadrilaterals have identical side lengths and angles measures. The second quadrilateral is rotated 90 degrees to the left. Wh

ich transformation maps quadrilateral EFGH to quadrilateral QRSP? dilation reflection rotation translation
Mathematics
2 answers:
mixas84 [53]3 years ago
4 0

Answer:

rotation

Step-by-step explanation:

elena-s [515]3 years ago
3 0

Answer:

C rotation

Step-by-step explanation:

edge 2020

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An isosceles trapazoid has base angles of 45 degrees and bases of lengths 9 and 15. The area of the trapazoid is what
lakkis [162]

Answer:

area of trapazoid= 1/2 × b× h

so = 1/2×9×15

=1/2×135

=67.5cm

6 0
2 years ago
Use the identity below to complete the tasks:
Gennadij [26K]

Answer:

a3+b3=(a+b)(a2-ab+b2)

4 0
3 years ago
What is the surface area
Drupady [299]
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8 0
3 years ago
Cot^2x csc^2x + 2 csc^2x –cot^2 = 2
kow [346]

Answer:

x = π/2 + πk

Step-by-step explanation:

cot² x csc² x + 2 csc² x − cot² x = 2

Multiply both sides by sin² x:

cot² x + 2 − cos² x = 2 sin² x

Add cos² x to both sides:

cot² x + 2 = 2 sin² x + cos² x

Pythagorean identity:

cot² x + 2 = sin² x + 1

Subtract 1 from both sides:

cot² x + 1 = sin² x

Pythagorean identity:

csc² x = sin² x

Multiply both sides by sin² x:

1 = sin⁴ x

Take the fourth root:

sin x = ±1

Solve for x:

x = π/2 + 2πk, 3π/2 + 2πk

Which simplifies to:

x = π/2 + πk

4 0
3 years ago
Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
3 years ago
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