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2 years ago

5

PLEASE HELP WILL GIVE BRAINLIEST DO NUMBER 10

1 answer:

Licemer1 [7]2 years ago

5 0

Answer:

<h2><u>369</u></h2>

Step-by-step explanation:

Since the equation of the line is y = 1/3x , or y = 0.33333333x , you multiply y, seltzer, by 3. You do this because x is 3 times the value of y.

Graphing the equation can be seen below

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The value of -9(square root) is not-3 because

love history [14]

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A) $(-3)^{2} \neq -9$

Step-by-step explanation:

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2 years ago

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Which represents the explicit formula for the arithmetic sequence an=7-3(n-1)

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Answer:

Hi how are you doing today Jasmine

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2 years ago

baherus [9]

Answer: see proof below

<u>Step-by-step explanation:</u>

Given: cos 330 = $\frac{\sqrt3}{2}$

Use the Double-Angle Identity: cos 2A = 2 cos² A - 1

$\text{Scratchwork:}\quad \bigg(\dfrac{\sqrt3 + 2}{2\sqrt2}\bigg)^2 = \dfrac{2\sqrt3 + 4}{8}$

Proof LHS → RHS:

LHS cos 165

Double-Angle: cos (2 · 165) = 2 cos² 165 - 1

⇒ cos 330 = 2 cos² 165 - 1

⇒ 2 cos² 165 = cos 330 + 1

Given: $2 \cos^2 165 = \dfrac{\sqrt3}{2} + 1$

$\rightarrow 2 \cos^2 165 = \dfrac{\sqrt3}{2} + \dfrac{2}{2}$

Divide by 2: $\cos^2 165 = \dfrac{\sqrt3+2}{4}$

$\rightarrow \cos^2 165 = \bigg(\dfrac{2}{2}\bigg)\dfrac{\sqrt3+2}{4}$

$\rightarrow \cos^2 165 = \dfrac{2\sqrt3+4}{8}$

Square root: $\sqrt{\cos^2 165} = \sqrt{\dfrac{4+2\sqrt3}{8}}$

Scratchwork: $\cos^2 165 = \bigg(\dfrac{\sqrt3+1}{2\sqrt2}\bigg)^2$

$\rightarrow \cos 165 = \pm \dfrac{\sqrt3+1}{2\sqrt2}$

Since cos 165 is in the 2nd Quadrant, the sign is NEGATIVE

$\rightarrow \cos 165 = - \dfrac{\sqrt3+1}{2\sqrt2}$

LHS = RHS $\checkmark$

4 0

3 years ago

PLEASE SHOW WORK

CaHeK987 [17]

(C)

Step-by-step explanation:

The volume of the conical pile is given by

$V = \dfrac{\pi}{3}r^2h$

Taking the derivative of V with respect to time, we get

$\dfrac{dV}{dt} = \dfrac{\pi}{3}\dfrac{d}{dt}(r^2h)$

$\:\:\:\:\:\:\:= \dfrac{\pi}{3}\left(2rh\dfrac{dr}{dt} + r^2\dfrac{dh}{dt}\right)$

Since r is always equal to h, we can set

$\dfrac{dr}{dt} = \dfrac{dh}{dt}$

so that our expression for dV/dt becomes

$\dfrac{dV}{dt} = \dfrac{\pi}{3}\left(3r^2\dfrac{dh}{dt}\right)$

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

Solving for dh/dt, we get

$\dfrac{dh}{dt} = \dfrac{1}{\pi r^2}\dfrac{dV}{dt}$

$\:\:\:\:\:\:\:= \dfrac{1}{9\pi\:\text{m}^2}(36\:\text{m}^3\text{/s})$

$\:\:\:\:\:\:\:= \dfrac{4}{\pi}\:\text{m/s}$

7 0

3 years ago