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2 years ago

I need help plzzzzzz

Mathematics

1 answer:

sammy [17]2 years ago

3 0

the answer is 0 Numble D

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20 POINTS AND BRAINLIEST!!!!!!!!!!!!!!!!!!

wolverine [178]

Look for points with whole number coefficients. (2,4) is such a point; (0,1) is another. To find points on the graph of the inverse of this function, all we have to do in this case is to reverse the order of the coefficients, so

(2,4) becomes (4,2) and (0,1) becomes (1,0). Only (4,2) is included in the list of answer choices, above, so the correct answer is (4,2).

7 0

3 years ago

You and some friends have $30. You want to order large

inna [77]

You can buy 3 pizzas and 3 drinks

5 0

3 years ago

PLEASE HELP ASAP OF YOU DO FOD BLESS YOU BECAUSE IT WILL RAISE MY GRADE

Sati [7]

Answer:

Step-by-step explanation:

The basic form of the equation is y=mx+b. The b stands for the y-intercept. In this equation, 8 is b due to its position in the equation, and -8/7 is m. Since we know that 8 is b and that b is the y-intercept, we know that 8 is the y-intercept.

Hope this helps!

4 0

1 year ago

Read 2 more answers

There were 276 people on an airplane. Write a number greater than 276

solniwko [45]

A number greater than 276 is 277.

6 0

3 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