Which day had the greatest increase in gas price? I’ll mark brainliest

Mathematics

1 answer:

Vesnalui [34]3 years ago

4 0

We need to find which has the greatest change in the slope and we can see it's friday as it's the most steepest.

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List the integers that satisfy both of these inequalities 2x+9<0 x>-12 put the answers on one line

lbvjy [14]

Answer:

Step-by-step explanation:

Given the inequality expressions

2x+9<0 and x>-12

2x+9<0

2x< 0-9

2x < -9

x < -9/2

x < -4.5

Since x >-12

-12 < x

Combining both inequalities

-12 <x < 4.5

<em>The integers between the intervals are -11, -10, -9, -8, -7, -6 and -5 </em>

6 0

3 years ago

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. Y = (4/9) x

frez [133]

Answer:

V = 8.06 cubed units

Step-by-step explanation:

You have the following curves:

$y_1=\frac{4}{9}x^2=f(x)\\\\y_2=\frac{13}{9}-x^2=g(x)$

In order to calculate the solid of revolution bounded by the previous curves and the x axis, you use the following formula:

$V=\pi \int_a^b [(g(x))^2-(f(x))^2]dx$ (1)

To determine the limits of the integral you equal both curves f=g and solve for x:

$f(x)=g(x)\\\\\frac{4}{9}x^2=\frac{13}{9}-x^2\\\\\frac{4}{9}x^2+x^2=\frac{13}{9}\\\\\frac{13}{9}x^2=\frac{13}{9}\\\\x=\pm 1$

Then, the limits are a = -1 and b = 1

You replace f(x), g(x), a and b in the equation (1):

$V=\pi \int_{-1}^{1}[(\frac{13}{9}-x^2)^2-(\frac{4}{9}x^2)^2]dx\\\\V=\pi \int_{-1}^1[\frac{169}{81}-\frac{26}{9}x^2+x^4-\frac{16}{81}x^4]dx\\\\V=\pi \int_{-1}^1 [\frac{169}{81}-\frac{26}{9}x^2+\frac{65}{81}x^4]dx\\\\V=\pi [\frac{169}{81}x-\frac{26}{27}x^3+\frac{65}{405}x^5]_{-1}^1\\\\V\approx8.06\ cubed\ units$