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2 years ago

HELP ME I will give brainliest to whoever is RIGhT

Mathematics

1 answer:

earnstyle [38]2 years ago

3 0

Answer:

Step-by-step explanation:

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Write a compound inequality to represent all of the numbers between -4 and 6.

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You use > for greater than and < for less then

-4 < x < 6

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3 years ago

Write an equation to describe the relationship between the independent variable x and the dependent variable y.

AlexFokin [52]

Answer:

x=y+2

Step-by-step explanation:

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3 years ago

Select the correct equation that describes the table below

matrenka [14]

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the answer is number 4......

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3 years ago

Solve |x – 4| + 6 = 17.

const2013 [10]

Answer:

$\mathrm{C)\:}x=15,\:-7$

Step-by-step explanation:

For $|a-b|=c$ , there are two possible cases:

$\begin{cases}a-b=c\\a-b=-c\\\end{cases}$

Solving for both cases, we get:

$\begin{cases}x-4+6=17, \:x+2=17, \: \fbox{$x=15$}\\-(x-4)+6=17,\: -x+10=17, \: -x=7, \:, \fbox{$x=-7$}\\\end{cases}$

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2 years ago

The volume V of an ice cream cone is given by V = 2 3 πR3 + 1 3 πR2h where R is the common radius of the spherical cap and the c

Nuetrik [128]

Answer:

The change in volume is estimated to be 17.20 $\rm{in^3}$

Step-by-step explanation:

The linearization or linear approximation of a function $f(x)$ is given by:

$f(x_0+dx) \approx f(x_0) + df(x)|_{x_0}$ where $df$ is the total differential of the function evaluated in the given point.

For the given function, the linearization is:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh$

Taking $R_0=1.5$ inches and $h=3$ inches and evaluating the partial derivatives we obtain:

$V(R_0+dR, h_0+dh) = V(R_0, h_0) + \frac{\partial V(R_0, h_0)}{\partial R}dR + \frac{\partial V(R_0, h_0)}{\partial h}dh\\V(R, h) = V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh$

substituting the values and taking $dx=0.1$ and $dh=0.3$ inches we have:

$V(R_0+dR, h_0+dh) =V(R_0, h_0) + (\frac{2 h \pi r}{3} + 2 \pi r^2)dR + (\frac{\pi r^2}{3} )dh\\V(1.5+0.1, 3+0.3) =V(1.5, 3) + (\frac{2 \cdot 3 \pi \cdot 1.5}{3} + 2 \pi 1.5^2)\cdot 0.1 + (\frac{\pi 1.5^2}{3} )\cdot 0.3\\V(1.5+0.1, 3+0.3) = 17.2002\\\boxed{V(1.5+0.1, 3+0.3) \approx 17.20}$

Therefore the change in volume is estimated to be 17.20 $\rm{in^3}$

4 0

3 years ago