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omeli [17]
3 years ago
5

Question 17/23

Mathematics
1 answer:
disa [49]3 years ago
6 0

Answer:

60%

Step-by-step explanation:

Formula:

Percentage decrease = (amount of decrease)/(original price)

✪ Solve ✪

3.85 - 1.54/3.85 x 100

2.31/3.85 x 100 = 60%

Hence, Percent decrease is 60%

<em>~Lenvy</em>

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What is the volume of a right circular cylinder with a radius of 5 cm and a height of 12 cm?
tatiyna

Answer:

V = 942.48cm³

Step-by-step explanation:

The volume of a right circular cylinder with a radius of 5 cm and a height of 12 cm is 942.48.

Formula: V=πr2h

V=πr2h=π·52·12≈942.4778cm³

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3 years ago
Fatima Sheroud sells children's clothing for The
tatyana61 [14]

Answer:

7890

Step-by-step explanation:

(5000*4%)+(x*5%)=594.50

200+(x*5%)=594.50

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6 0
4 years ago
Plz help with full process
BabaBlast [244]

Answer:

\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1

Step-by-step explanation:

<u>Algebra Simplifying</u>

We are given the expression:

\displaystyle T=\frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }

We need to repeatedly use the following identity:

(a^2-b^2)=(a-b)(a+b)

For example, the first numerator has a difference of squares thus we factor as:

(a - b)^{2} - {c}^{2} =(a-b-c)(a-b+c)

The first denominator can be factored also:

{a}^{2} - {(b+ c)}^{2} = (a-b-c)(a+b+c)

Applying the same procedure to all the expressions:

\displaystyle T=\frac{(a - b- c)(a-b+c) }{a-b-c)(a+b+c) } + \frac{(b - c - a)(b-c+a) }{(b-c-a)(b+c+a) } + \frac{(c - a-b)(c-a+b) }{(c-a-b)(c+a+b)}

Simplifying all the fractions:

\displaystyle T=\frac{a - b- c}{a+b+c} + \frac{b-c+a}{b+c+a } + \frac{c-a+b }{c+a+b}

Since all the denominators are equal:

\displaystyle T=\frac{a - b+ c+b-c+a+c-a+b}{a+b+c}

Simplifying:

\displaystyle T=\frac{a +c+b}{a+b+c}

Simplifying again:

T = 1

Thus:

\boxed{\displaystyle \frac{(a - b)^{2} - {c}^{2} }{{a}^{2} - {(b+ c)}^{2} } + \frac{(b - c)^{2} - {a}^{2} }{{b}^{2} - {(c+ a)}^{2} } + \frac{(c - a)^{2} - {b}^{2} }{{c}^{2} - {(a+ b)}^{2} }=1}

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