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3 years ago

one teaspoon equals 0.5 centimeters. how many liters equals 50 teaspoons? Round the answer to the nearest hundredth.

Mathematics

2 answers:

alukav5142 [94]3 years ago

8 0

50 teaspoons would equal 50*0.5 = 25 centimeters

1 cm = 0.001 liter

25*0.001 = 0.025 liters

Karo-lina-s [1.5K]3 years ago

5 0

50 teaspoons would equal 50*0.5 = 25 centimeters

1 cm = 0.001 liter

25*0.001 = 0.025 liters

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What single percentage change is equivalent to a 11% decrease followed by a 13% decrease?

Sunny_sXe [5.5K]

Answer: its 22.57%

Step-by-step explanation:

I got it wrong on hegarty so I could show you the answer since your an army lol

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3 years ago

Can the sides of a triangle have lengths 13, 5, and 16?

Alika [10]

Answer:

Yes

Step-by-step explanation:

13 + 5 > 16
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2 years ago

Point K is on line segment JL. Given JK = 5x + 7, JL = 2x+8, and KL = 4,

kati45 [8]

Answer:

The numerical length of JL is 6 units

Step-by-step explanation:

Here, we want to determine the numerical length of JL

Mathematically;

JL = JK + KL

2x + 8 = 5x + 7 + 4

2x + 8 = 5x + 11

5x -2x = 8-11

3x = -3

x = -1

But JL = 2x + 8

JL = 2(-1) + 8 = -2 + 8 = 6

3 0

3 years ago

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

3 years ago

Please help !! <br> basic triangle proofs

jeka57 [31]

Answer:

AD=AC (as D is midpoint to from line AC the line AD=AC

Step-by-step explanation:

HENCE PROVED

8 0

3 years ago