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2 years ago

7

4. Spencer and Sydney both have many books at home. Together they have 65 books. Spencer has 9 fewer books than Sydney. How many

books does Spencer have? 28 O 74 27 O 37

1 answer:

AlekseyPX2 years ago

7 0

Answer:

they are both reading book;

Step-by-step explanation:

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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

laiz [17]

In general, the volume

$V=\pi r^2h$

has total derivative

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

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.

8 0

2 years ago

4. Which of the following lines is perpendicular to y = -2X + 8?

Alenkinab [10]

For this case we have to by definition, if two lines are perpendicular then the product of its slopes is -1.

That is to say:

$m_ {1} * m_ {2} = - 1$

We have the following equation:

$y = -2x + 8$

So:

$m_ {1} = - 2$

Thus:

$m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {- 2}\\m = \frac {1}{2}$

Thus, a line perpendicular to the given line must have slope $m = \frac {1} {2}.$

Option A:

$x + 2y = 8\\2y = -x + 8\\y = - \frac {1} {2} x + 4$

It is not perpendicular!

Option B:

$x-2y = 6\\2y = x-6\\y = \frac {1} {2} x-3$

If it is perpendicular!

Option C:

$2x + y = 4\\y = -2x + 4$

It is not perpendicular!

Option D:

$2x-y = 1\\y = 2x-1$

It is not perpendicular!

The correct option is option B

ANswer:

Option B

5 0

3 years ago

How do I do proofs?

Neko [114]

Is this algebra? What type of Math this

7 0

2 years ago

Please help. I’ll mark you as brainliest if correct!!!!

aliya0001 [1]

Answer:

(-∞, -6] U [-2, ∞)

Step-by-step explanation:

To solve this, begin by factoring this quadratic equation into its factored form:

x² + 8x + 12 ≥ 0 becomes (x+6)(x+2) ≥ 0.

x = -6, and x = -2 are the zeros of this parabola. Therefore:

(-∞, -6] U [-2, ∞) are the parts of the graph above y = 0 because the graph

opens upward.

** Remember, when the '≥' sign is present, the

square brackets must be used.

8 0

2 years ago

Slove the simoltanous equation by using substitution or elimination method 5x-3y=1 3x+2y=1<br>

xxTIMURxx [149]

<u>By substitution method </u>

5x - 3y =1

3x + 2y = 1

_____o____o____

$5x - 3y = 1 \\ 5x = 1 + 3y \\ x = \frac{1 + 3y}{5}$

______o____o_____

$3x + 2y = 1 \\ 3( \frac{1 + 3y}{5} ) + 2y = 1 \\ \frac{3 + 9y}{5} + 2y = 1 \\ \frac{3 +9y + 10y}{5} = 1 \\ \frac{3 + 19y}{5} = 1 \\ 3 + 19y = (1)(5) \\ 3 + 19y = 5 \\ 19y = 5 - 3 \\ 19y = 2 \\ y = \frac{2}{19}$

______o_____o____

$x = \frac{1 + 3y}{5} = \frac{1 + 3 \frac{2}{19} }{5} \\ x = \frac{1 + \frac{6}{19} }{5} = \frac{ \frac{19 + 6}{19} }{5} = \frac{ \frac{25}{19} }{5} \\ x = \frac{25}{19} \div 5 \\ x = \frac{25}{19} \times \frac{1}{5} \\ x = \frac{5}{19}$

I hope I helped you^_^

7 0

2 years ago