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3 years ago

8

ritu can row downstream 20km in 2 hrs and upstream 4km in 2 hrs. find the speed of rowing in still water and the speed of the cu

rrent

1 answer:

Marina CMI [18]3 years ago

7 0

Answer:

Speed of boat in still water is 6km/h

Speed of current is 4km/h

Step-by-step explanation:

We know speed = distance / time

Let the speed of boat in still water be x

Let the speed of current be y

Hence the speed of boat, down stream = $x+y$

Distance downstream = 20km

Time taken = 2 hours

We have

$x+y=\frac{20}{2}$

$x+y=10$ ......(1)

Speed of boat upstream = $x-y$

Distance upstream = 4

Time taken = 2

We have

$x-y=\frac{4}{2}$

$x-y=2$ ...(2)

Adding (1) & (2)

$2x=12$

$x=6$

Substituting in (1) we get

$y=4$

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Suppose a point at (2, 3) is translated to (7, −1). Which rule describes this translation?

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3 years ago

Suppose the half-life of a material is 10 days. You have one 1 kg of the material today. How much of the material would you have

boyakko [2]

Answer:

0.50 kg of the material would be left after 10 days.

0.25 kg of the material would be left after 20 days.

Step-by-step explanation:

We have been given that the half-life of a material is 10 days. You have one 1 kg of the material today. We are asked to find the amount of material left after 10 days and 20 days, respectively.

We will use half life formula.

$A=a\cdot(\frac{1}{2})^{\frac{t}{h}}$ , where,

A = Amount left after t units of time,

a = Initial amount,

t = Time,

h = Half-life.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{10}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{1}$

$A=1\text{ kg}\cdot\frac{1}{2}$

$A=0.5\text{ kg}$

Therefore, amount of the material left after 10 days would be 0.5 kg.

$A=1\text{ kg}\cdot(\frac{1}{2})^{\frac{20}{10}}$

$A=1\text{ kg}\cdot(\frac{1}{2})^{2}$

$A=1\text{ kg}\cdot\frac{1^2}{2^2}$

$A=1\text{ kg}\cdot\frac{1}{4}$

$A=0.25\text{ kg}$

Therefore, amount of the material left after 20 days would be 0.25 kg.

6 0

3 years ago

rachel is frosting the top of two cakes and wants to know which cake will require more frosting. one cake is round with a diamet

Jet001 [13]

Answer:

The square cake is has a larger area by about 40.73 square inches

Step-by-step explanation:

Find the area of the top of each cake.

The square cake has an area of 81 sq inches, ( length x width = Area, the

length is 9)

The round cake has an area of 16π, which is about 50.27 square inches

(A = πr², here r is 4, radius is half the diameter)

81 - 16π = 81 - 50.27 = 40.73 square inches

4 0

2 years ago

Which of the following are solutions to the equation below? Check all that apply. x2 + x - 20 = 0

Mila [183]

Δ = 1 + 4*20 = 81
x' = (-1 + 9)/2 = 4
x'' = (-1 - 9)/2 = -5
4² + 4 - 20 = 0
(-5)² - 5 - 20 = 0
Ok !

7 0

2 years ago

Read 2 more answers

KL = 4x +3.<br> JK = 4x + 2, and<br> JL = 45<br> Find KL.

lbvjy [14]

Answer:

$KL=23$

Step-by-step explanation:

We are given the length of the whole line and its two halves. Since JK and KL make up the line JL, adding them together would make the line JL.

$(4x+3)+(4x+2) = 45$

$8x+5=45$

$8x=40$

$x=5$

Now we can plug the value of $x$ back into the expression we were given for the line KL.

$4(5)+3$

$23$

7 0

2 years ago