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klio [65]
3 years ago
12

Carl went to the store to buy items for his party.He bought two packs for plates for $3.24 and a pack of cups for $5.70. He also

bought a pack of forks for $6.17. If Carl paid with two $10 bills ,how much change did he get back
Mathematics
1 answer:
Andre45 [30]3 years ago
7 0

3.24 + 5.70 + 6.17 = 15.11

20.00 - 15.11 = 5.11

He got 5.11$ of change back.

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In the figure, solve for x and y.
evablogger [386]

Answer:

y=124°

x=67°

Step-by-step explanation:

For the triangle on the right

180-74-50=56°

then y= 180-56=124°

For the triangle on the middle (that small triangle)

180-85(Head opposite angles)-50=45°

So

x=180-45-68=67°

5 0
3 years ago
Dana invests £5000 for 4 years in a savings account She gets 2% per annum compound interest in the first year, then x% for 3 yea
Vanyuwa [196]

Answer:

£1,330.46

Hope this helps, mate!

Step-by-step explanation:

Hope this helps, mate! :)

6 0
2 years ago
5x 20 and the. multiplicative inverse of 5
ehidna [41]
That is 5*20 is 100 and the inverse of 5 is 25
8 0
3 years ago
Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y
DiKsa [7]

The area of the surface is given exactly by the integral,

\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx

We have

y(x)=\dfrac15x^5\implies y'(x)=x^4

so the area is

\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the i-th subinterval are, respectively,

\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2

r_i=\dfrac{5-0}{10}i=\dfrac i2

with midpoint

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

with 1\le i\le10.

Over each subinterval, we interpolate f(x)=\sqrt{1+x^8} with the quadratic polynomial,

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

Then

\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that the latter integral reduces significantly to

\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)

which is about 651.918, so that the area is approximately 651.918\pi\approx\boxed{2048}.

Compare this to actual value of the integral, which is closer to 1967.

4 0
3 years ago
A problem states: "There are 9 more children than parents in a room. There are 25 people in the room in all. How many children a
zepelin [54]

Answer:

c-9

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
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