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3 years ago

Which expression shows the prime factorization of 36?

1 answer:

6 0

36....we need to break this down...12 * 3 = 36
12 * 3....we can break this down because 12 is not prime...4 * 3 = 12
4 * 3 * 3...we can break the 4 down because it is not prime
2 * 2 * 3 * 3 <==== prime factorization

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Simplifying radicals: √196s²

aliya0001 [1]

√196s² = √196 times √s²

s² is the square of 's'
196 is the square of 14
So both can easily come out of the radical.

√196s² = <u>14s</u>

3 0

3 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

2 years ago

What is the value of the expression 3y−zy+5z3y−zy+5z when y = 6 and z = 2?

WARRIOR [948]

The valve of the expression would be 184

6 0

3 years ago

Read 2 more answers

3x~6y=~3<br> 2x+4y=30<br> Elimination using multiplication

Nat2105 [25]

$\left \{ {{3x-6y=-3\ \ |*2} \atop {2x+4y=30\ \ | *(-3)}} \right. \\\\ \left \{ {{6x-12y=-6} \atop {-6x-12y=-90}} \right. \\+----\\elimination\\\\
0 \neq -69\\\\There\ are\ no\ solutions$

7 0

2 years ago

Please help with the equation

Anastaziya [24]

Answer:

The distance is dependent

Step-by-step explanation:

Time is always the dependent variable because time will always stay the same when other things change, such as distance.

3 0

3 years ago