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3 years ago

HELP ME PLEASE ASAP!!

Mathematics

1 answer:

timama [110]3 years ago

8 0

Answer:

Step-by-step explanation:

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Find the range for f(x)=8x-3,given the domain is {-2,4,8,10}

Setler [38]

Answer:

-19, 29, 61, 77

Step-by-step explanation:

Put the domain numbers as a solution for x. May or may not be right. Sorry if it's wrong...

6 0

3 years ago

Read 2 more answers

5 small containers of capacity 16, 72, 12, 24 and 56 Litres are to be used to fill a bigger container. What is the capacity of t

PIT_PIT [208]

Answer: 1008

Step-by-step explanation:

Given

5 small containers of capacity 16,72,12,24, and 56 litres is required to fill a big container.

The volume of big container is the LCM of the 5 given containers

$LCM(16,72,12,24,56)=1008$

Therefore, the capacity of the bigger container is 1008 litres.

6 0

2 years ago

Help me answer this please

nikitadnepr [17]

Answer:

x=115 y=65

Step-by-step explanation:

You get 115 for X because angle x is vertical to 115 degrees.

You get 65 because y is supplementary to x, and 180-65=115.

4 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

Order the decimals from least to greatest.<br><br> 1.45, 2.56, 0.99

zvonat [6]

Answer:

0.99, 1.45 then 2.56 will be the greatest!

7 0

3 years ago