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4 years ago

What is the 6th row of Pascal's triangle?

Mathematics

1 answer:

Grace [21]4 years ago

8 0

Answer:

1, 6, 15, 20, 15, 6, 1

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What is the area of a Hexagon with a radius of 5 in?

lukranit [14]

75 in^2, because 5 times 5 (the split up hexagons make parallelograms and there is 3 of them, and parallelogram area is base times height.)

7 0

3 years ago

The question is below thanks

Katyanochek1 [597]

this is ur complete Venn diagram

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3 years ago

If 70 and above is considered a passing grade, how many students passed the math test?

Snowcat [4.5K]

<h3>Answer: D) 8</h3>

Explanation:

The green, light blue, and red bars all indicate scores that are 70 or higher.

We add up the heights of those bars to find out how many people passed.

green bar = height of 2
light blue bar = height of 5
red bar = height of 1

So 2+5+1 = 8 people passed.

Unfortunately 2 people (dark blue bar) got below 70 so they didn't pass.

3 0

2 years ago

Read 2 more answers

Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ

SOVA2 [1]

$y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}$

The arc length of the curve is

$\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx$

which has a value of about 5.99086.

Let $f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}$ . Split up the interval of integration into 10 subintervals,

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

The left and right endpoints are given respectively by the sequences,

$\ell_i=\dfrac{i-1}2$

$r_i=\dfrac i2$

with $1\le i\le10$ .

These subintervals have midpoints given by

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

Over each subinterval, we approximate $f(x)$ with the quadratic polynomial

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

so that the integral we want to find can be estimated as

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

It turns out that

$\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{f(\ell_i)+4f(m_i)+f(r_i)}6$

so that the arc length is approximately

$\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086$

5 0

3 years ago

I need help with q6 please

choli [55]

This is 25 divided by 2000 times 100

6 0

3 years ago