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MAVERICK [17]
3 years ago
14

Can I get help please

Mathematics
1 answer:
Arisa [49]3 years ago
6 0
Miles is going to spend fewer hours bicycling this week than he did last year.
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Please help with this pleaseee
saw5 [17]

Answer:

CED=126

Step-by-step explanation:

The total of all 3 angles should be 180

so:

180-38-16 (ABC is equal to ECD)

180-38-16=126

126=CED

5 0
2 years ago
Read 2 more answers
The estimated value of the integral from 0 to 2 of x cubed dx , using the trapezoidal rule with 4 trapezoids is
bulgar [2K]
The integral is approximated by the sum,

\displaystyle\int_0^2f(x)\,\mathrm dx\approx\sum_{n=0}^4\frac12\times\frac{f(x_n)+f(x_{n+1})}2=\frac14\sum_{n=0}^3(f(x_n)+f(x_{n+1}))

where f(x)=x^3 and x_n=\dfrac12n, giving you

\displaystyle\frac14\sum_{n=0}^3\bigg(\left(\frac n2\right)^3+\left(\frac{n+1}2\right)^3\bigg)
\displaystyle\frac1{32}\sum_{n=0}^3(n^3+(n+1)^3)
\displaystyle\frac1{32}\sum_{n=0}^3(2n^3+3n^2+3n+1)

Faulhaber's formulas make short work of computing the sum. You have

\displaystyle\sum_{n=0}^k1=k+1
\displaystyle\sum_{n=0}^kn=\frac{k(k+1)}2
\displaystyle\sum_{n=0}^kn^2=\frac{k(k+1)(2k+1)}6
\displaystyle\sum_{n=0}^kn^3=\frac{k^2(k+1)^2}4

which gives

\displaystyle\frac1{16}\sum_{n=0}^3n^3+\frac3{32}\sum_{n=0}^3n^2+\frac3{32}\sum_{n=0}^3n+\frac1{32}\sum_{n=0}^31
\displaystyle\frac{36}{16}+\frac{42}{32}+\frac{18}{32}+\frac4{32}
\implies\displaystyle\int_0^2x^3\,\mathrm dx\approx\frac{17}4=4.25
4 0
3 years ago
How would u location 8/3 and -11/3 on a number line?
tia_tia [17]

Answer:

it would be point A if there is a point A

8 0
2 years ago
Solve the system by substitution.<br> -5x + 3y = 2<br> y = 2x
hammer [34]

Answer:

Step-by-step explanation:

-5x +3(2x) = 2

-5x +6x = 2

x=2

4 0
3 years ago
Best answer will be branliest and win 10 points! Please helpppp
Veseljchak [2.6K]
350mL because it has more for a cheaper price
6 0
3 years ago
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