1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Leviafan [203]
3 years ago
11

What is 2÷1/6 in simplest form?

Mathematics
1 answer:
Alexandra [31]3 years ago
3 0
2 ÷ ( 1 / 6 ) = 2 x 6 = 12 ;
You might be interested in
Whats 7 times 8 divided by 2 i think the answer s 6 am i right or ring please tell me
slamgirl [31]

Answer:

<h2><em><u>28</u></em></h2>

Step-by-step explanation:

<em><u>First</u></em><em><u>,</u></em>

7 times 8 = 7 × 8 = 56

<em><u>Then</u></em><em><u>,</u></em>

The product divided by 2 = 56 ÷ 2 = 28

<em><u>Hence</u></em><em><u>,</u></em>

<em><u>The</u></em><em><u> </u></em><em><u>required</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>28</u></em>

5 0
2 years ago
1/2-3/16 hurry pls cause I really need it don’t judge me
damaskus [11]
The answer is 2/16 if you do the math with the numerators only.
7 0
2 years ago
Graph h(x)=-x^4 +3
Katyanochek1 [597]
Hope this helps, the y intercept is 3 ans the x intercepts are +1.3 and -1.3

8 0
3 years ago
Sanjeev had 350 marbles and he gave 1/5 of them to his friend, Rayson. How many
kicyunya [14]

Answer:

70

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Please help meeeee math​
Sergeeva-Olga [200]

Q2. By the chain rule,

\dfrac{dy}{dx} = \dfrac{dy}{dt} \cdot \dfrac{dt}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}

We have

x=2t \implies \dfrac{dx}{dt}=2

y=t^4+1 \implies \dfrac{dy}{dt}=4t^3

The slope of the tangent line to the curve at t=1 is then

\dfrac{dy}{dx} \bigg|_{t=1} = \dfrac{4t^3}{2} \bigg|_{t=1} = 2t^3\bigg|_{t=1} = 2

so the slope of the normal line is -\frac12. When t=1, we have

x\bigg|_{t=1} = 2t\bigg|_{t=1} = 2

y\bigg|_{t=1} = (t^4+1)\bigg|_{t=1} = 2

so the curve passes through (2, 2). Using the point-slope formula for a line, the equation of the normal line is

y - 2 = -\dfrac12 (x - 2) \implies y = -\dfrac12 x + 3

Q3. Differentiating with the product, power, and chain rules, we have

y = x(x+1)^{1/2} \implies \dfrac{dy}{dx} = \dfrac{3x+2}{2\sqrt{x+1}} \implies \dfrac{dy}{dx}\bigg|_{x=3} = \dfrac{11}4

The derivative vanishes when

\dfrac{3x+2}{2\sqrt{x+1}} = 0 \implies 3x+2=0 \implies x = -\dfrac23

Q4. Differentiating  with the product and chain rules, we have

y = (2x+1)e^{-2x} \implies \dfrac{dy}{dx} = -4xe^{-2x}

The stationary points occur where the derivative is zero.

-4xe^{-2x} = 0 \implies x = 0

at which point we have

y = (2x+1)e^{-2x} \bigg|_{x=0} = 1

so the stationary point has coordinates (0, 1). By its "nature", I assume the question is asking what kind of local extremum this point. Compute the second derivative and evaluate it at x=0.

\dfrac{d^2y}{dx^2}\bigg|_{x=0} = (8x-4)e^{-2x}\bigg|_{x=0} = -4 < 0

The negative sign tells us this stationary point is a local maximum.

Q5. Differentiating the volume equation implicitly with respect to t, we have

V = \dfrac{4\pi}3 r^3 \implies \dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}

When r=5\,\rm cm, and given it changes at a rate \frac{dr}{dt}=-1.5\frac{\rm cm}{\rm s}, we have

\dfrac{dV}{dt} = 4\pi (5\,\mathrm{cm})^2 \left(-1.5\dfrac{\rm cm}{\rm s}\right) = -150\pi \dfrac{\rm cm^3}{\rm s}

Q6. Given that V=400\pi\,\rm cm^3 is fixed, we have

V = \pi r^2h \implies h = \dfrac{400\pi}{\pi r^2} = \dfrac{400}{r^2}

Substitute this into the area equation to make it dependent only on r.

A = \pi r^2 + 2\pi r \left(\dfrac{400}{r^2}\right) = \pi r^2 + \dfrac{800\pi}r

Find the critical points of A.

\dfrac{dA}{dr} = 2\pi r - \dfrac{800\pi}{r^2} = 0 \implies r = \dfrac{400}{r^2} \implies r^3 = 400 \implies r = 2\sqrt[3]{50}

Check the sign of the second derivative at this radius to confirm it's a local minimum (sign should be positive).

\dfrac{d^2A}{dr^2}\bigg|_{r=2\sqrt[3]{50}} = \left(2\pi + \dfrac{1600\pi}{r^3}\right)\bigg|_{r=2\sqrt[3]{50}} = 6\pi > 0

Hence the minimum surface area is

A\bigg_{r=2\sqrt[3]{50}\,\rm cm} = \left(\pi r^2 + \dfrac{800\pi}r\right)\bigg|_{r=2\sqrt[3]{50}\,\rm cm} = 60\pi\sqrt[3]{20}\,\rm cm^2

Q7. The volume of the box is

V = 8x^2

(note that the coefficient 8 is measured in cm) while its surface area is

A = 2x^2 + 12x

(there are two x-by-x faces and four 8-by-x faces; again, the coefficient 12 has units of cm).

When A = 210\,\rm cm^2, we have

210 = 2x^2 + 12x \implies x^2 + 6x - 105 = 0 \implies x = -3 \pm\sqrt{114}

This has to be a positive length, so we have x=\sqrt{114}-3\,\rm cm.

Given that \frac{dx}{dt}=0.05\frac{\rm cm}{\rm s}, differentiate the volume and surface area equations with respect to t.

\dfrac{dV}{dt} = (16\,\mathrm{cm})x \dfrac{dx}{dt} = (16\,\mathrm{cm})(\sqrt{114}-3\,\mathrm{cm})\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{4(\sqrt{114}-3)}5 \dfrac{\rm cm^3}{\rm s}

\dfrac{dA}{dt} = 4x\dfrac{dx}{dt} + (12\,\mathrm{cm})\dfrac{dx}{dt} = \left(4(\sqrt{114}-3\,\mathrm {cm}) + 12\,\mathrm{cm}\right)\left(0.05\dfrac{\rm cm}{\rm s}\right) = \dfrac{\sqrt{114}}5 \dfrac{\rm cm^2}{\rm s}

5 0
1 year ago
Other questions:
  • Sue has 4 pieces of wood. The lengths of her pieces of wood are 1/3 foot, 2/5 foot, 3/10, and 1/4 foot. Which piece of wood is t
    14·1 answer
  • Solve the equation using the distributive property and properties of equality. -5(a+3)=-55 What is the value of a? A.)-14 B.) -8
    6·1 answer
  • 12. The square and rectangle shown below have the same perimeter length Show that the length of the rectangle (3x + 1) centimetr
    15·1 answer
  • Find the LCM= Least common multiple for these two sets of numbers. ​
    12·1 answer
  • What is 5x23x2 using any property
    8·1 answer
  • Last question for the night.
    9·2 answers
  • Which of the following is an ordered pair on the graph? What does it represent in this situation?
    8·1 answer
  • The slope of the line below is 2. Use the coordinates of the labeled point to find a point-slope equation of the line.
    9·1 answer
  • Point J is located at (-6, -2) on the coordinate plane. Point J is reflected over the
    9·1 answer
  • Write these polynomials in standard form<br> 1). 5x^2- 15+7<br> 2.) 4 + 8n- 3n^4+ 10n2
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!