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d1i1m1o1n [39]
3 years ago
6

Answer this pls asap

Mathematics
2 answers:
Elena-2011 [213]3 years ago
6 0

Step-by-step explanation:

(2x-11)°+(6x-7)°=180°(being sum of the angle of the triangle is 180°

=2x-11+6x-7=180°

=8x+4=180°

=8x=180-4

=8x=176

=x=176/8

=x=22

Scorpion4ik [409]3 years ago
5 0

Answer:

24.75

Step-by-step explanation:

2x-11 + 6x-7 =180

8x-18=180

8x=198 /8

x=24.75

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WILL GIVE BRANLIEST (15 POINTS)
Strike441 [17]

Answer:

x^2 + 2x+1

Step-by-step explanation:

g(x) = x^2

Replace x with x+1

g(x+1) = (x+1)^2 = (x+1)(x+1)

FOIL

g(x+1) = x^2+x+x+1 = x^2 + 2x+1

5 0
3 years ago
20 points!<br> I need help with these!
defon

1. 113.64

(work) 947x.12

2. 130.5

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Alex Ar [27]
2x^2 + 2y^2 + 2xy I think
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3 years ago
consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where
boyakko [2]

Answer:

a)X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

b)Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

Step-by-step explanation:

From the question we are told that

The Function

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

Generally the differentiation of function f(x) is mathematically solved as

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

f(x)=\frac{x^3+x^2+5x+1}{x^2}

Therefore

f'(x)=\frac{x^2+10x+3}{x^4}

Generally critical point is given as

f'(x)=0

\frac{x^2+10x+3}{x^4}=0

x=-5 \pm\sqrt{22}

Generally the maximum and minimum x value for critical point is mathematically solved as

f'(-5 \pm\sqrt{22})

Where

Maximum value of x

f'(-5 +\sqrt{22})

Minimum value of x

f'(-5 +\sqrt{22})

Therefore interval of increase is mathematically given by

f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})

f(x)

Therefore interval of decrease is mathematically given by

(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)

Generally the second differentiation of function f(x) is mathematically solved as

f''(x)=\frac{2(x^2+15x+6)}{x^5}

Generally the point of inflection is mathematically solved as

f''(x)=0

x^2+15x+6=0

Therefore inflection points is given as

x=\frac{1}{2} (-15 \pm \sqrt{201}

f''(x)>0,\frac{1}{2}(-15-\sqrt{201})

a)Generally the concave upward interval X is mathematically given as

X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

f''(x)

b)Generally the concave downward interval Y is mathematically given as

Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

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B. Only Sequence B

The one with the reflection over the line HI and then the 6 unit translation to the right
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