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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
Mathematics
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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UkoKoshka [18]

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

3x + 180 = 4x

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

3x + 3y = 1200

-

3x – 5y = - 200

=

8y = 1400

Y = 175

X = 225

Kelly = 225

Nelly = 175

Learn More:

brainly.com/question/2273184

brainly.com/question/11876335

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3 0
3 years ago
Read 2 more answers
Calculus 2
FinnZ [79.3K]

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:

\displaystyle \Delta x=\frac{4-1}{6}=\frac{1}{2}

Therefore, the trapezoidal approximation is:

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

Evaluate:

\displaystyle =\frac{1}{4}(7)(\sqrt{\ln(1)}+2\sqrt{\ln(1.5)}+...+2\sqrt{\ln(3.5)}+\sqrt{\ln(4)})\\\\\approx18.139337

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

Thus:

\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:

\displaystyle =\frac{1}{2}(7)\Big(f(1.25)+f(1.75)+...+f(3.25)+f(3.75)\Big)\\\\ =\frac{1}{2}(7) (\sqrt{\ln(1.25)}+\sqrt{\ln(1.75)}+...+\sqrt{\ln(3.25)}+\sqrt{\ln(3.75)})\\\\\approx 18.767319

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:

\displaystyle =\frac{1}{6}(7)(\sqrt{\ln(1)}+4\sqrt{\ln(1.5)}+2\sqrt{\ln(2)}+4\sqrt{\ln(2.5)}+...+4\sqrt{\ln(3.5)}+\sqrt{\ln(4)})\\\\\approx 18.423834

6 0
2 years ago
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