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3 years ago

A ratio that is equivalent to 60:125

1 answer:

6 0

There are multiple other choices.
1.) 12:25.
2.) (As a fraction) 12/25.

For the comment u made for 60:132.
1.) 30:66
2.) 15:33
3.) 5:11

Those are all the equivalent ratios of those numbers.

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a ladder leans against the sufe of a house. the angle of elevation of the ladder is 70 degrees when the bottom of the ladder is

nekit [7.7K]

Answer:

≈ 35.1 ft

Step-by-step explanation:

The model is a right triangle with ladder being the hypotenuse and the angle between the ground and the ladder is 70°

Using the cosine ratio, with l being the length of the ladder.

cos70° = $\frac{adjacent}{hypotenuse}$ = $\frac{12}{l}$ ( multiply both sides by l )

l × cos70° = 12 ( divide both sides by cos70° )

l = $\frac{12}{cos70}$ ≈ 35.1 ( to the nearest tenth )

The ladder is approx 35.1 ft long

4 0

2 years ago

*Click the photo*

Charra [1.4K]

Answer:

137.5 Miles

Step-by-step explanation:

2 = 50 Right? So if you were to divide 50 by 2, you would get 1 = 25. All you would have to do from there would be to multiply 25 times 5.5 to get 137.5.

8 0

3 years ago

Read 2 more answers

YOOO PLEASE HELPPPP!!!!

Evgesh-ka [11]

Given:

Base length of triangle = 40 units

Height of triangle = 9 units

Length of hypotenuse of triangle = 41 units

To find:

Find the value of Tan A

Steps:

Tan of an angle is equal to the opposite length by adjacent length.

So,

Tan A = $\frac{opposite}{adjacent}$

Tan A = $\frac{9}{40}$

Tan A = 0.225

Therefore, the exact value of Tan A is 0.225.

Happy to help :)

If you need any help, feel free to ask

4 0

2 years ago

Please help me with another problem

lorasvet [3.4K]

"one" will fill both blanks.

8 0

3 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

2 years ago