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3 years ago

Calculate 1/3 of 50 cm Plssss helpppp solve....

Mathematics

2 answers:

Westkost [7]3 years ago

8 0

1/3 x 50= 50/3 = 16 2/3= 16.33

Lelechka [254]3 years ago

3 0

Answer:

Step-by-step explanation:

16.6

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What is the domain of the relation shown? will mark as branliest

Schach [20]

Answer:

2 or 4

Step-by-step explanation:

6 0

3 years ago

Please help. 10 points and brainliest.

Snezhnost [94]

Answer:

$\frac{a^2}{b^2}$

Step-by-step explanation:

Apply the exponent rule $a^b*a^c=a^b+c$ to a

$a^4*a^-^2=a^4^-^2=a^2$

$b^3a^2b^-^5$

Apply the exponent rule to b

$b^3*b^-^5=b^3^-^5=b^-^2$

Apply the exponent rule $a^-^b=\frac{1}{a^b}$

$b^-^2=\frac{1}{b^2}=\frac{1}{b^2}a^2=\frac{a^2}{b^2}$

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3 years ago

Air is being pumped into a spherical balloon at a rate of 5 cm^3/min. Determine the rate at which the radius of the balloon is i

Romashka-Z-Leto [24]

0.08 cm/min

Step-by-step explanation:

Given:

$\dfrac{dV}{dt}=5\:\text{cm}^3\text{/min}$

Find $\frac{dr}{dt}$ when diameter D = 20 cm.

We know that the volume of a sphere is given by

$V = \dfrac{4\pi}{3}r^3$

Taking the time derivative of V, we get

$\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt} = 4\pi\left(\dfrac{D}{2}\right)^2\dfrac{dr}{dt} = \pi D^2\dfrac{dr}{dt}$

Solving for $\frac{dr}{dt}$ , we get

$\dfrac{dr}{dt} = \left(\dfrac{1}{\pi D^2}\right)\dfrac{dV}{dt} = \dfrac{1}{\pi(20\:\text{cm}^2)}(5\:\text{cm}^3\text{/min})$

$\:\:\:\:\:\:\:= 0.08\:\text{cm/min}$

3 0

3 years ago

Help I don't get it and I need it done fast.take your time if your going to help.

Zigmanuir [339]

Answer:

Step-by-step explanation:

4 0

3 years ago

Someone help me fast please

lions [1.4K]

Answer:

Step-by-step explanation:

Perimeter of WXY = WSY+WRX+XY

--> WSY = SY x 2

--> WSY = 16 x 2 = 32

Since it is an isosceles triangle, WRX = WSY

--> WRX = 32

--> Draw a straight line from W to XY to divide it into two halves assuming it to be point A. This would form a right angle triangle of WAX.

--> Solve it using the cos theta rule

--> Angle = Angle X = 70°

Hypotenuse = WRX = 32

Adjacent = WA = ?

--> Cos (Angle) = Adjacent/Hypotenuse

Cos (70) = WA/32

WA = 10.9 rounded off to 11

--> WA=AY= 11

--> XY = WA + AY = 11+11 = 22

--> Perimeter = WSY+WRX+XY

Perimeter = 32+32+22

Perimeter = 86

Therefore, the perimeter of WXY is 86.

4 0

3 years ago