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miss Akunina [59]

2 years ago

6

Liam bought a sandwich for $4.75, a bag of chips for $2.30, and a drink for $1.15. The tax was $0.65. He gave the cashier $10.00

. How much change should Liam have receive

1 answer:

Nat2105 [25]2 years ago

5 0

Answer:
$1.15

step-by-step explanation:
4.75 + 2.30 + 1.15 + 0.65
8.85
10.00 - 8.85
1.15

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Answer:

(1)

Donna = Pounds 180

Kyra = Pounds 240

(2)

Kelly = Pounds 225

Nelly = Pounds 175

Step-by-step explanation:

(1)

Donna = x

Kyra = x + 60

If Donna spends 1/3 of her money she is left with 2/3 of her money;

2/3 x

Kyra now has twice as much money as Donna can be represented as;

X + 60 = 2 * 2/3 x

X + 60 = 4/3 x

(x + 60)3 = 4x

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180 = 4x – 3x

180 = x

Donna = 180

Kyra = 180 + 60 = 240

(2)

Kelly = x

Ned = y

Kelly spends 40% of her money. Therefore he is left with 60%;

60/100 x

Nelly spends Euros 40 of his money. Therefore he is left with;

y- 40

If they now have the same amount of money;

60/100 x = y – 40

3/5 x = y – 40

3x = 5y – 200

3x – 5y = -200

Remember Kelly & Ned has Euros 400 in total;

x + y = 400

Now we can solve the simultaneous equation by substitution;

x + y= 400

3x – 5y = -200

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-

3x – 5y = - 200

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Y = 175

X = 225

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Answer:

See Below.

Step-by-step explanation:

We want to estimate the definite integral:

$\displaystyle \int_1^47\sqrt{\ln(x)}\, dx$

Using the Trapezoidal Rule, Midpoint Rule, and Simpson's Rule with six equal subdivisions.

1)

The trapezoidal rule is given by:

$\displaystyle \int_{a}^bf(x)\, dx\approx\frac{\Delta x}{2}\Big(f(x_0)+2f(x_1)+...+2f(x_{n-1})+f(x_n)\Big)$

Our limits of integration are from x = 1 to x = 4. With six equal subdivisions, each subdivision will measure:

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2)

The midpoint rule is given by:

$\displaystyle \int_a^bf(x)\, dx\approx\sum_{i=1}^nf\Big(\frac{x_{i-1}+x_i}{2}\Big)\Delta x$

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$\displaystyle =\frac{1}{2}\Big(f\Big(\frac{1+1.5}{2}\Big)+f\Big(\frac{1.5+2}{2}\Big)+...+f\Big(\frac{3+3.5}{2}\Big)+f\Big(\frac{3.5+4}{2}\Big)\Big)$

Simplify:

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3)

Simpson's Rule is given by:

$\displaystyle \int_a^b f(x)\, dx\approx\frac{\Delta x}{3}\Big(f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+...+4f(x_{n-1})+f(x_n)\Big)$

So:

$\displaystyle =\frac{1/2}{3}\Big((f(1)+4f(1.5)+2f(2)+4f(2.5)+...+4f(3.5)+f(4)\Big)$

Simplify:

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2 years ago