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3 years ago

7

140 fluid oz increased by 30%

1 answer:

gladu [14]3 years ago

4 0

So 130% of 140 fluid ounce? 182 fl oz.

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A certain radioactive isotope decays at a rate of 2% per 100 years. if t represents time in years and y represents the amount of

gregori [183]

We rewrite the equation:
y = y0 * e ^ (- 0.0002 * t)
In this equation:
I = represents the initial amount of the isotope
We have then:
89% of the isotope left:
0.89 * y0 = y0 * e ^ (- 0.0002 * t)
We clear the time:
e ^ (- 0.0002 * t) = 0.89
Ln (e ^ (- 0.0002 * t)) = Ln (0.89)
-0.0002 * t = Ln (0.89)
t = Ln (0.89) / (- 0.0002)
t = 582.6690813
round to the nearest year:
t = 583 years
Answer:
There will be 89% of the isotope left in about:
t = 583 years

7 0

3 years ago

The price of an item has increased 15% since last year. However, a person can buy the item for a 25% employee discount. The empl

ANEK [815]

Answer: $200

Explanation:

The employee pays $172.50 and he got 25% employee discount.

If the cost is $100, he pays $(100 - 25) = $75.

So, employee pays =

100

75

⋅

172.50

= $230.00.

Again, price was increased by 15% last year. So whose cost is $100

is sold by $(100+15) = $115.

When sell price is $115, then cost price is $100. Then,

when sell price is $230, then cost price is

100

115

⋅

230

= $200.

5 0

3 years ago

A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

7 0

3 years ago

Math plz help it's hard (8th grade math)

Mrac [35]

This is what I got I dont know if It should be an equation but that's all i can get to, gl

8 0

3 years ago

Read 2 more answers

Angela is going on vacation to Mexico. She needs to buy new shirts for her trip. She buys 10 new shirts that cost her a total of

vivado [14]

Angela purchased 4 casual shirts and 6 dressy shirts.

$25 × 6 = $150
$18 × 4 = $72

$150 + $72 = $222

hope this helped :)

8 0

3 years ago