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3 years ago

15

What is the speed of the ball if it travels 19km in 19min

1 answer:

kifflom [539]3 years ago

8 0

Speed=distance/time

s=19km/19min=1km/min

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zhannawk [14.2K]

Answer:

Your correct answer is B. If the price of the television was 31/5. the price of the laptop was $532

Step-by-step explanation:

You can divide it on a calculatoror on paper and get it correct! Hope it helped! Please mark brainliest! Have a wonderful day!!!

8 0

3 years ago

A solid cylinder of radius 7cm is 10cm long.find it's total surface area

Anton [14]

Answer:

C. 238π cm^3.

Step-by-step explanation:

Surface area = 2 * area of the circular ends + area of the curved length

= 2π r^2 + 2π r* l

= 2 π 7^2 + 2π 7 * 10

= 98π + 140π

= 238π cm^3.

7 0

3 years ago

Read 2 more answers

There are 100 bacteria in a jar. Every day, the number of bacteria increases 25%. In approximately how many days will there be 5

algol [13]

Use compound interest formula

$\\ \rm\Rrightarrow P(1+r/100)^t=500$

$\\ \rm\Rrightarrow 100(1+0.25)^t=500$

$\\ \rm\Rrightarrow (1.25)^t=5$

$\\ \rm\Rrightarrow log(1.25)^t=log5$

$\\ \rm\Rrightarrow tlog1.25=log5$

$\\ \rm\Rrightarrow t(0.097)=0.3$

$\\ \rm\Rrightarrow t=3.3\approx 3$

Option A

5 0

2 years ago

Read 2 more answers

The team's combined score was 0. What was Travis's score

Pavlova-9 [17]

Answer:

Sorry but we need more info.

Step-by-step explanation:

:)

8 0

3 years ago

Solve the given differential equation by using an appropriate substitution. The DE is a Bernoulli equation.

Mama L [17]

Multiplying both sides by $y^2$ gives

$xy^2\dfrac{\mathrm dy}{\mathrm dx}+y^3=1$

so that substituting $v=y^3$ and hence $\frac{\mathrm dv}{\mathrm dv}=3y^2\frac{\mathrm dy}{\mathrm dx}$ gives the linear ODE,

$\dfrac x3\dfrac{\mathrm dv}{\mathrm dx}+v=1$

Now multiply both sides by $3x^2$ to get

$x^3\dfrac{\mathrm dv}{\mathrm dx}+3x^2v=3x^2$

so that the left side condenses into the derivative of a product.

$\dfrac{\mathrm d}{\mathrm dx}[x^3v]=3x^2$

Integrate both sides, then solve for $v$ , then for $y$ :

$x^3v=\displaystyle\int3x^2\,\mathrm dx$

$x^3v=x^3+C$

$v=1+\dfrac C{x^3}$

$y^3=1+\dfrac C{x^3}$

$\boxed{y=\sqrt[3]{1+\dfrac C{x^3}}}$

6 0

3 years ago