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3 years ago

A cone has a diameter of 4 cm and a height of 11 cm. What is the volume 1 point

Mathematics

1 answer:

anzhelika [568]3 years ago

4 0

Answer: 46.05

Step-by-step explanation:

1/3 * 3.14 * 4 * 11 = 46.05 (rounded to the tenth)

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13. Find the equations of the straight lines which passes

ipn [44]

9514 1404 393

Answer:

y = 1/2x +2
y = -2x +7

Step-by-step explanation:

The slope of a line is the tangent of the angle it makes with the x-axis. The given line has a slope of -1/3, so the lines we want will have slopes of ...

m1 = tan(arctan(-1/3) +45°) = 0.5 . . . . . using a calculator

m2 = tan(arctan(-1/3) -45°) = -2

Of course, these two lines are perpendicular to each other, so their slopes will have a product of -1: (0.5)(-2) = -1.

We can use the point-slope form of the equation for a line to write the desired equations:

y = m(x -h) +k . . . . . line with slope m through point (h, k)

Line 1:

y = 1/2(x -2) +3

y = 1/2x +2

Line 2:

y = -2(x -2) +3

y = -2x +7

3 0

2 years ago

What is 2x+3=9?????????????

ale4655 [162]

Answer:

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

2 years ago

If the technician charged $327.50, then the technician worked ___ hours.

Anika [276]

How much was he paid per hour

8 0

2 years ago

Read 2 more answers

Help please asap with this

Allushta [10]

Use photo math app it will help you

The second one 1.7 = 17/10
-0.048 = -6/125
98% = 98/100= 49/50

5 0

2 years ago