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3 years ago

Solve for x: You must show your work (how your got your answer). 8x + 5 = 35

1 answer:

Rudiy273 years ago

4 0

8x + 5 = 35
------------------
Subtract 5 from each side
8x + 5 - 5 = 35 - 5
8x = 30
-----------------------------
Divide each side by the number 8
8x ÷ 8 = 30 ÷ 8
x = 15/4 or 3.75 or 3 3/4

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I need help with a math question that is 0.1x = 5/y

BabaBlast [244]

Unfortunately you have not included the directions. Are you supposed to solve for x? or for y?

Suppose the directions say, "Solve for x." Then, clear out the decimal fraction by mult. both sides of the given equation by 10. This will give you the solution, that is, a formula for x in terms of y.

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3 years ago

109/10 in simplest form explain

maksim [4K]

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3 years ago

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lutik1710 [3]

Pythagorean Theorem: a^2 + b^2 = c^2

-A and B are legs of the triangle

-C is the hypotenuse

#1.

6^2 + 8^2 = c^2

36 + 64 = c^2

100 = c^2

c = 10

#2.

5^2 + 12^2 = c^2

25 + 144 = c^2

169 = c^2

c = 13

#4.

10^2 + 4^2 = c^2

100 + 16 = c^2

116 = c^2

c = $4\sqrt{29}$

#5.

15^2 + 9^2 = c^2

225 + 81 = c^2

306 = c^2

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#6.

120^2 + b^2 = 150^2

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#7.

144^2 + b^2 = 194^2

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Hope this helps!! :)

4 0

3 years ago

1) What is the value of the expression 3 4 + 15.45 - 19.75 + 7.87?

Delvig [45]

Answer:

37.57

Step-by-step explanation:

34 + 15.45 - 19.75 + 7.87

34 + 15.45 = 49.45

49.45 - 19.75 + 7.87

49.45 + 7.87 = 57.32

57.32 - 19.75 = 37.57

6 0

2 years ago

Define the double factorial of n, denoted n!!, as follows:n!!={1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n} if n is odd{2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n} if n is evenand (

tekilochka [14]

Answer:

Radius of convergence of power series is $\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{1}{108}$

Step-by-step explanation:

Given that:

n!! = 1⋅3⋅5⋅⋅⋅⋅(n−2)⋅n n is odd

n!! = 2⋅4⋅6⋅⋅⋅⋅(n−2)⋅n n is even

(-1)!! = 0!! = 1

We have to find the radius of convergence of power series:

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

Power series centered at x = a is:

$\sum_{n=1}^{\infty}c_{n}(x-a)^{n}$

$\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}](8x+6)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}n!(3n+3)!(2n)!!}{2^{n}[(n+9)!]^{3}(4n+3)!!}]2^{n}(4x+3)^{n}\\\\\sum_{n=1}^{\infty}[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}](x+\frac{3}{4})^{n}\\$

$a_{n}=[\frac{8^{n}4^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}n!(3(n+1)+3)!(2(n+1))!!}{[(n+1+9)!]^{3}(4(n+1)+3)!!}]\\\\a_{n+1}=[\frac{8^{n+1}4^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]$

Applying the ratio test:

$\frac{a_{n}}{a_{n+1}}=\frac{[\frac{32^{n}n!(3n+3)!(2n)!!}{[(n+9)!]^{3}(4n+3)!!}]}{[\frac{32^{n+1}(n+1)!(3n+6)!(2n+2)!!}{[(n+10)!]^{3}(4n+7)!!}]}$

$\frac{a_{n}}{a_{n+1}}=\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

Applying n → ∞

$\lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}= \lim_{n \to \infty}\frac{(n+10)^{3}(4n+7)(4n+5)}{32(n+1)(3n+4)(3n+5)(3n+6)+(2n+2)}$

The numerator as well denominator of $\frac{a_{n}}{a_{n+1}}$ are polynomials of fifth degree with leading coefficients:

$(1^{3})(4)(4)=16\\(32)(1)(3)(3)(3)(2)=1728\\ \lim_{n \to \infty}\frac{a_{n}}{a_{n+1}}=\frac{16}{1728}=\frac{1}{108}$

4 0

3 years ago