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3 years ago

How much lashonda paid ?

Mathematics

1 answer:

Nonamiya [84]3 years ago

7 0

61 - (61 x .85) = 9.15

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Given the information in the picture, which trigonometric identity can be used to solve for the height of the blue ladder that i

Stells [14]

The trigonometric identity can be used to solve for the height of the blue ladder that is leaning against the building is $\rm sin=\dfrac{o}{h}$ .

We have to determine

Which trigonometric identity can be used to solve for the height of the blue ladder that is leaning against the building?.

<h3>Trigonometric identity</h3>

Trigonometric Identities are the equalities that involve trigonometry functions and hold true for all the values of variables given in the equation.

Trig ratios help us calculate side lengths and interior angles of right triangles:

The trigonometric identity that can be used to solve for the height of the blue ladder is;

$\rm Sin47=\dfrac{50}{H}\\\\H=\dfrac{50}{sin47}\\\\H=68 feet$

Hence, the trigonometric identity can be used to solve for the height of the blue ladder that is leaning against the building is $\rm sin=\dfrac{o}{h}$ .

To know more about trigonometric identity click the link given below.

brainly.com/question/1256744

6 0

2 years ago

⎧

d1i1m1o1n [39]

<u>Given</u>:

The given expression to find the nth term of the sequence is $d(n)=d(n-1) \cdot (-5)$

The first term of the sequence is $d(1)=8$

We need to determine the third term of the sequence.

<u>Second term:</u>

The second term of the sequence can be determined by substituting n = 2 in the nth term of the sequence.

Thus, we have;

$d(2)=d(2-1) \cdot (-5)$

$d(2)=d(1) \cdot (-5)$

$d(2)=8 \cdot (-5)$

$d(2)=-40$

Thus, the second term of the sequence is -40.

<u>Third term:</u>

The third term of the sequence can be determined by substituting n = 3 in the nth term of the sequence.

Thus, we have;

$d(3)=d(3-1) \cdot (-5)$

$d(3)=-40 \cdot (-5)$

$d(3)=120$

Thus, the third term of the sequence is 120.

7 0

3 years ago

Read 2 more answers

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

2 years ago

Read 2 more answers

Help me please! If you DO decide to help, please explain it to me since I'm extremely confused! THANK YOU!

den301095 [7]

$\bf \cfrac{(-4x^2)(2x^{-2}y)^3}{(16x^5)(4y^3)^2}\implies \cfrac{(-4x^2)(2^3x^{-2\cdot 3}y^3)}{(16x^5)(4^2y^{3\cdot 2})}\implies \cfrac{(-4x^2)(8x^{-6}y^3)}{(16x^5)(16y^6)} \\\\\\ \cfrac{-32x^{2-6}y^3}{256x^5y^6}\implies -\cfrac{x^{-4} y^3}{8x^5y^6}\implies -\cfrac{1}{8x^5x^{4}y^6y^{-3}}\implies -\cfrac{1}{8x^{5+4}y^{6-3}} \\\\\\ -\cfrac{1}{8x^9y^3}$

8 0

3 years ago

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Which of the following statements is TRUE? a. If your original creditor (ex: credit card company) sells your debt to a collectio

nikitadnepr [17]

Answer:

the answer is d

Step-by-step explanation:

i just know its fine.

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3 years ago

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