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3 years ago

15

A store pays $3 for a shirt they sell it for $5.70 what is the percent markup

1 answer:

REY [17]3 years ago

7 0

Store pays ($3)
Store sells it for ($5.70)
The percent is always (100%) when there selling it.

markup= $3 ÷ $5.70 · 100% = 1.9

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Cost to store: $65<br> Markup:25%<br> Selling price:?

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100% + 25% is 120% then 120 as a decimal 1.20 then you do 1.20 x 65 = 78
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3 years ago

2 lb, 891g, 1 kg, 0.02 T greatest to least

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To arrange in descending order or greatest to least, we will first convert all the values in same unit.

Lets convert all the values in kg

1 lb = 0.45 kg

2 lb = $2\times0.45=0.9$ kg

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891 g = $891\times0.001=0.891$ kg

1 T = 907.185 kg

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0.02T, 1 kg, 2 lb, 891 g

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A triangle has two sides of lengths 7 and 9. What value could the length of the third side be?

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8 and 10 are possible lengths of the third side.

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U-substitutions only work for specific kinds of expressions. Below, you are asked to choose a value of n for which u-substitutio

Dahasolnce [82]

Answer:

Step-by-step explanation:

(a) $\int x^n e^{5x^4+1} \ dx$

Suppose $5x^4 + 1 = f$

by differentiation;

$\implies \ 20 x^3 dx = df --- (1)$

Suppose n = 3

Then, the integral

$I = \int x^ 3 e^{5x^4 + 1} \ dx$

$= \int e^f \ \dfrac{df}{20}$

$= \dfrac{1}{20} \int e^f \ dt$

$= \dfrac{1}{20} e^f + C$

recall that $f = 5x^4 + 1$

Then;

$\mathbf{ I = \dfrac{1}{20}e^{5x^4+1}+C}$

(b) $\int \dfrac{cos (\dfrac{1}{x^3})}{x^n } \ dx$

suppose; $\dfrac{1}{x^3} = f$

$x^3 = f$

$\implies -3x^{-4} \ dx = df$

$\implies \dfrac{1}{x^4} \ dx =-\dfrac{1}{3} df$

If n = u, then the integration is:

$I = \int \dfrac{1}{x^4} \ cos (\dfrac{1}{x^4}) \ dx$

$= \int -\dfrac{1}{3} \ cos \ f \ df$

$= -\dfrac{1}{3} \int \ cos \ f \ df$

$= -\dfrac{1}{3} \ sin \ f + C$

Since; $x^3 = f$

Then;

$\mathbf {I = -\dfrac{1}{3} \ sin \ \Big( \dfrac{1}{x^3}\Big) + C}$

(c) $\int \dfrac{x+n}{x^2 + 8x -4} \ dx$

Suppose $x^2 + 8x - 4 = f$

Then, by differentiation of both sides

$(2x + 8) \ dx = df$

$(x + 4) \ dx = \dfrac{1}{2} \ df$

Suppose n = 4 in integration, then:

$I = \int \dfrac{(x + 4) }{x^2 +8x -4} \ dx$

By substitution;

$I = \int \dfrac{1}{2}\dfrac{1}{f} \ df$

$= \dfrac{1}{2} \ \ { In |f|} + C$

$\mathbf{= \dfrac{1}{2} \ \ { In |x^2+8x -4|} + C}$

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3 years ago