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drek231 [11]
3 years ago
11

Which expressions are equivalent to (5g+3h+4)\cdot2(5g+3h+4)⋅2left parenthesis, 5, g, plus, 3, h, plus, 4, right parenthesis, do

t, 2 ?
Choose all answers that apply:

Choose all answers that apply
A
(5g+3h)\cdot8(5g+3h)⋅8left parenthesis, 5, g, plus, 3, h, right parenthesis, dot, 8

(Choice B)
B
(5g+3h)\cdot6(5g+3h)⋅6left parenthesis, 5, g, plus, 3, h, right parenthesis, dot, 6

(Choice C)
C
None of the above
Mathematics
2 answers:
lions [1.4K]3 years ago
7 0

Answer:

C. None of the above

Semenov [28]3 years ago
6 0

Answer:

C None of the above

Step-by-step explanation:

The expression

(5g+3h+4)⋅2

can be expanded using distributive property as follows:

5g⋅2 + 3h⋅2 + 4⋅2  =

= 10g + 6h + 8

option A expression

(5g+3h)⋅8

can be expanded using distributive property as follows:

5g⋅8+3h⋅8 =

= 40g + 24h

which is different from 10g + 6h + 8

Option B expression

(5g+3h)⋅6

can be expanded using distributive property as follows:

5g⋅6+3h⋅6 =

= 30g + 18h

which is different from 10g + 6h + 8

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What is the perimeter of the triangle shown on the coordinate plane, to the nearest tenth of a unit?
seraphim [82]

21.6 units

Calculate the length of the 2 inclined sides using the distance formula

d = √(x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = (- 3, 3 ) and (x₂, y₂ ) = (3, 4 )

d = √(3 + 3 )² + (4 - 3)² = √(36 + 1 ) = √37

repeat for (x₁, y₁ ) = (- 3, 3 ) and (x₂, y₂ ) = (3, - 3 )

d = √(3 + 3 )² + (- 3 - 3 )² = √(36 + 36 ) = √72

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3 0
3 years ago
A remote-control airplane is descending at a rate of
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83 m

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3 years ago
Find the taylor polynomial t3(x) for the function f centered at the number
inysia [295]

Answer:

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

Step-by-step explanation:

We are given that

f(x)=7tan^{-1}(x)

a=1

T_n(x)=\sum_{r=0}^{n}\frac{f^r(a)(x-a)^r}{r!}

Substitute n=3 and a=1

t_3(x)=f(1)+f'(1)(x-1)+\frac{f''(1)(x-1)^2}{2!}+\frac{f'''(1)(x-1)^3}{3!}

f(x)=7tan^{-1}(x)

f(1)=7tan^{-1}(1)=7\times \frac{\pi}{4}=\frac{7\pi}{4}

Where tan^{-1}(1)=\frac{\pi}{4}

f'(x)=\frac{7}{1+x^2}

Using the formula

\frac{d(tan^{-1}(x))}{dx}=\frac{1}{1+x^2}

f'(1)=\frac{7}{2}

f''(x)=\frac{-14x}{(1+x^2)^2}

f''(1)=-\frac{7}{2}

f''(x)=-14x(x^2+1)^{-2}

f'''(x)=-14((x^2+1)^{-2}-4x^2(x^2+1)^{-3}})

By using the formula

(uv)'=u'v+v'u

f'''(x)=-14(\frac{x^2+1-4x^2}{(1+x^2)^3}

f'''(x)=(-14)\frac{-3x^2+1}{(1+x^2)^3}

f'''(1)=-14(\frac{-3(1)+1}{2^3})=\frac{7}{2}

Substitute the values

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{2\times 3\times 2\times 1}(x-1)^3

t_3(x)=\frac{7\pi}{4}+\frac{7}{2}(x-1)-\frac{7}{4}(x-1)^2+\frac{7}{12}(x-1)^3

7 0
3 years ago
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