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3 years ago

Write the rule for the table below help me pleaseeeeeeeeee

Mathematics

1 answer:

Anna71 [15]3 years ago

7 0

-5 like minus 5 no mater the negetive for y
Hope this helped ")

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For all functions of the form f(x) = ax2 + bx + c, which is true when b = 0?

Klio2033 [76]

When b=0
f(x)=ax^2+c

test each
A. with x^2-1, A is false
B. with -x^2-1, B is false
C. cannot find contradiction
D. the axis is actually x=0, 0 is nithere positive nor negative, false

answer is C

7 0

4 years ago

Read 2 more answers

Solve the expression for x: (4 – 2) (x – 2) = 4x – 8 A. x = –2 B. x = 0 C. x = 6/5 D. x = 2

Ratling [72]

Answer:

Option D) x = 2

Step-by-step explanation:

(4-2)(x-2) = 4x - 8

Distribute and simplify 4-2

2(x-2) = 4x - 8

2x-4 = 4x -8

Subtract 2x on both sides and add 8 on both sies

4x - 8 = 2x-4

2x= 4

Divide 2 on both sides

x = 2

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7 0

4 years ago

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

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)$