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3 years ago

15

Dimtry bought a pair of pants at the discounted price of $30. The original price of the pants was $40. What was the percent of t

he discount? PLEASE EXPLAIN TO ME HOW YOU GOT YOUR ANSWER

1 answer:

AVprozaik [17]3 years ago

5 0

Answer:

25%

Step-by-step explanation:

Say we start from the beginning with $40. If the pants are discounted but we don't know the percent discount, then we need a variable, x:

40 - x% * 40 = 30

The total is 40. When we take a discount off of 40, we are taking a percent of 40 and subtracting that from 40. From the problem, this should be 30. Thus, the equation above.

Add x%*40 to both sides and subtract 30 from both sides:

x%*40 = 10

Divide by 40:

x% = 1/4 = 0.25

Multiply each side by 100:

x = 25

The answer is 25%.

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Lelechka [254]

Answer:

77. $\cot^{6} x = \cot^{4} x \csc^{2}x - \cot^{4} x$ Proved

78. $\sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ]$ Proved

79. $\cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x$ Proved.

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Step-by-step explanation:

77. Left hand side

= $\cot^{6} x$

= $\cot^{4} x \times \cot^{2} x$

= $\cot^{4}x [\csc^{2}x - 1]$

{Since we know, $\csc^{2} x - \cot^{2}x = 1$ }

= $\cot^{4} x \csc^{2}x - \cot^{4} x$

= Right hand side (Proved)

78. Left hand side

= $\sec^{4}x \tan^{2} x$

= $\sec^{2} x [1 + \tan^{2}x] \tan^{2} x$

{Since $\sec^{2}x - \tan^{2}x = 1$ }

= $\sec^{2}x [\tan^{2}x + \tan^{4}x ]$

= Right hand side (Proved)

79. Left hand side

= $\cos^{3} x\sin^{2} x$

= $\cos x[1 - \sin^{2} x] \sin^{2} x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $[\sin^{2}x - \sin^{4}x] \cos x$

= Right hand side

80. Left hand side

= $\sin^{4}x - \cos^{4}x$

= $[\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x$

{Since $\sin^{2}x + \cos^{2} x = 1$ }

= $1 - 2\cos^{2} x[1 - \cos^{2}x ]$

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3 years ago

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Natali5045456 [20]

Answer:

Perimeter = 180

Area = 1080

Step-by-step explanation:

Perimeter is length of all sides added together. 30+78+72 = 180

Area is the length times the height/2 for a triangle. 30*72/2 = 1080

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3 years ago

Al oeste de Albuquerque, Nuevo México, la carretera 40 con rumbo al este es recta y tiene una fuerte pendiente hacia la ciudad.

alexgriva [62]

Usando el concepto de pendiente, se encuentra que el cambio en la distancia horizontal es de 10 pies.

La pendiente de una reta es la <u>razón entre el cambio vertical y el cambio horizontal,</u> o sea, cual el cambio vertical cuando el cambio horizontal es de 1.

En este problema, hay que:

La pendiente es -100, o sea, cuando hay un desplazamiento de un pie por adelante, hay una queda de 100 pies.
Bajado 1000 pies, entonces, aplicando la proporción:

1 pie H -> -100 pies V

x pie H -> -1000 pies V

Aplicando la multiplicación cruzada:

$-100x = -1000$

$100x = 1000$

$x = \frac{1000}{100}$

$x = 10$

El cambio en la distancia horizontal es de 10 pies.

Un problema similar es dado en brainly.com/question/24766917

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Ksenya-84 [330]

Answer:

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We can combine like terms: $4\sqrt{3}+22\sqrt{3}=26\sqrt{3}$

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