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3 years ago

Click on all the points that are solutions to y = x^2 - 9.

Mathematics

2 answers:

NemiM [27]3 years ago

7 0

Answer:b and C

Step-by-step explanation:

Ymorist [56]3 years ago

5 0

A & B .........................

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What is absolute value

jeka94

Absolute value is the distance between a point and 0. Absolute value is always positive. When you're writing numbers in absolute value, make sure you add:

| |

The absolute value of a negative number is also positive. For example:

| -7 | = 7 ⇒The absolute value of -7 is 7.

3 0

3 years ago

Find the surface area of the cylinder in terms of pi. 9cm and 19cm

Natali [406]

Answer:

Surface area of cylinder= 211.5π

Step-by-step explanation:

Given,

Height of cylinder = 19cm

Diameter of cylinder = 9cm

i.e Radius of cylinder = diameter/2 = 9/2 = 4.5cm

Surface area of cylinder = 2πrh + 2πr²

Plug corresponding values of r and h in surface are formula.

Surface area of cylinder = 2πrh + 2πr² = 2π*4.5*19 + 2π*4.5²

Surface area of cylinder = 171π+40.5π = 211.5π

Hence we get surface area of cylinder= 211.5π

3 0

3 years ago

If sin theta =1/4, theta in quadrant II, find the exact value of cos[theta +pi/6]

Mashcka [7]

Using cos addition formula:
use x for theta
cos(x+π/6)=cosx*cos(π/6)-sinx*sin(π/6)
sinx=1/4
cosx=√15/4
cos(π/6)=√3/2
sin(π/6)=1/2
cos(x+π/6)=(√15/4*√3/2)-(1/4*1/2)
cos(x+π/6)=(√45/8)-(1/8 )
cos(x+π/6)=(√45-1)/8)

6 0

3 years ago

Solve this!

Reptile [31]

Withdraw- remove
Deposit- put
Discount- take off

5 0

2 years ago

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

2 years ago