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4 years ago

Assuming the lines are parallel angel 4 corresponds with angle ___

Mathematics

2 answers:

Soloha48 [4]4 years ago

6 0

Answer:

∠ 2

Step-by-step explanation:

∠ 2 corresponds to ∠ 4 because each is in the same position, the upper right- hand corner, in its group of 4 angles.

nydimaria [60]4 years ago

5 0

Answer:

Angle 2

Step-by-step explanation:

They're in the same positions on the same transversal and different parallel lines. Stay safe :)

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The values for three sets of data are shown below.

soldi70 [24.7K]

Answer:

<h2><em><u>nearest tenth of a mile. The data is listed below. Compute the range, standard deviation and variance of the data. 1.1 5.2 3.6 5.0 4.8 1.8 2.2 5.2 1.5 0.8 40) compare three data sets. Without calculating, determine which data set has the greatest sample standard deviation and which has the least sample standard deviation</u></em></h2>

Step-by-step explanation:

brainlist i did show work plzzzz

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3 years ago

Read 2 more answers

Mass-35g<br> Density-8.80g/cm3<br> What is volume?

GuDViN [60]

You should down load cheggstudy

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3 years ago

This year there are 205 students in seventh grade. Last year there were 3% less

alina1380 [7]

205×3/100=6.15

Total students

205-6.15=198.85

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198.89=199 students

5 0

3 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

Help!! Please thanks

Juli2301 [7.4K]

Answer:

i think you have chosen the right answer i don't really know its tricky for me

Step-by-step explanation:

3 0

3 years ago