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2 years ago

7

cert 22.2 U To 3. Draw an area diagram to find (0.36) (0.53). Label and organize your work so that it can be followed by others.

1 answer:

Masteriza [31]2 years ago

3 0

The result of the product of (0.36) and (0.53) is 0.1908

<h3>How to determine the product</h3>

The product expression is given as:

(0.36) * (0.53)

Rewrite the product as follows:

$(0.36) * (0.53) = (0.3 + 0.06) * (0.5 + 0.03)$

Expand the product

$(0.36) * (0.53) = 0.3 * (0.5 + 0.03)+ 0.06 * (0.5 + 0.03)$

Expand

$(0.36) * (0.53) = 0.15 + 0.009+ 0.03+ 0.0018$

Evaluate the sum

$(0.36) * (0.53) = 0.1908$

See attachment for the area diagram of the product

Read more about products at:

brainly.com/question/10873737

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3 years ago

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Answer:

Part 1) "a" value is $2$

Part 2) The vertex is the point $(1,8)$

Part 3) The equation of the axis of symmetry is $x=1$

Part 4) The vertex is a minimum

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Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

$y=a(x-h)^{2}+k$

where

(h,k) is the vertex of the parabola

if a > 0 then the parabola open upward (vertex is a minimum)

if a < 0 then the parabola open downward (vertex is a maximum)

The equation of the axis of symmetry of a vertical parabola is equal to the x-coordinate of the vertex

so

$x=h$

In this problem we have

$y=2(x-1)^{2}+8$ -----> this is the equation in vertex form of a vertical parabola

The value of $a=2$

so

a>0 then the parabola open upward (vertex is a minimum)

The vertex is the point $(1,8)$

so

$(h,k)=(1,8)$

The equation of the axis of symmetry is $x=1$

The equation of a vertical parabola in standard form is equal to

$y=ax^{2}+bx+c$

Convert vertex form in standard form

$y=2(x-1)^{2}+8$

$y=2(x^{2}-2x+1)+8$

$y=2x^{2}-4x+2+8$

$y=2x^{2}-4x+10$

see the attached figure to better understand the problem

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3 years ago

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3 years ago

the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const

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In general, the volume

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$\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)$

If the cylinder's height is kept constant, then $\dfrac{\mathrm dh}{\mathrm dt}=0$ and we have

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}$

which is to say, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ are directly proportional by a factor equivalent to the lateral surface area of the cylinder ( $2\pi r h$ ).

Meanwhile, if the cylinder's radius is kept fixed, then

$\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

since $\dfrac{\mathrm dr}{\mathrm dt}=0$ . In other words, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ are directly proportional by a factor of the surface area of the cylinder's circular face ( $\pi r^2$ ).

Finally, the general case ( $r$ and $h$ not constant), you can see from the total derivative that $\dfrac{\mathrm dV}{\mathrm dt}$ is affected by both $\dfrac{\mathrm dh}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ in combination.

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3 years ago