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3 years ago

Find the gcf of the monomial factor 15a-25b+20

Mathematics

1 answer:

Olin [163]3 years ago

6 0

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0=4t-16t^2 Solve this please

vaieri [72.5K]

Answer:

$\large\boxed{t=0\ \vee\ t=\dfrac{1}{4}}$

Step-by-step explanation:

$4t-16t^2=0\qquad\text{divide both sides by 4}\\\\\dfrac{4t}{4}-\dfrac{16t^2}{4}=\dfrac{0}{4}\\\\t-4t^2=0\\\\t(1-4t)=0\iff t=0\ \vee\ 1-4t=0\\\\1-4t=0\qquad\text{subtract 1 from both sides}\\\\-4t=-1\qquad\text{divide both sides by (-4)}\\\\t=\dfrac{1}{4}$

7 0

3 years ago

How is using base ten blocks helps you to add decimals

lapo4ka [179]

It helps to add decimals because it means one out of ten

7 0

3 years ago

Read 2 more answers

Which is greater 5/6 or 2/3

geniusboy [140]

5/6 because 5/6=0.80 and 2/3=0.66

6 0

3 years ago

Read 2 more answers

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

Describe a situation in which the rate of change for data over a short period of time would be a good estimate for the rate of c

aalyn [17]

Answer:

will you let me off the hook because of my cute pika pfp?

5 0

2 years ago