All categories

4 years ago

If you love your mother 76.5% more than you love your father how much do you love your father?

Mathematics

1 answer:

Yakvenalex [24]4 years ago

4 0

Well 100% is a whole so you would subtract 76.5% from 100% and get your answer.

You might be interested in

A candle maker uses a mold to make candles like

Mashutka [201]

Answer: 340 cm3

Step-by-step explanation:

Hi, to answer this question we have to apply the next formulas:

Circumference = 2π radius

Replacing with the circumference value and solving for r:

31.416 =2π r

31.416 /2π =r

5cm =r

And then:

Volume of a cone = 1/3 x π x radius^2 x height

Replacing with the values given:

V = 1/3 π (5)^2 (13) = 340 cm3

Feel free to ask for more if needed or if you did not understand something.

4 0

4 years ago

Let c = 10.Which expression makes the comparison true.

den301095 [7]

Answer:

c² - 63

Step-by-step explanation:

4c - 10 = 30

3c ÷ 3 = 10

c² - 63 = 37

c² ÷ 10 = 10

Since c² - 63 = 37 and c + 25 = 35, 37 is greater than 35 so it would make sense to choose c² - 63 as the answer.

c + 25 < 37

I hope this helps you :D

3 0

3 years ago

Which number is written in scientific notation? A. 6x10^11 B. 12x10^-8 C. 0.2x10^5 D. 8x7^2

Lesechka [4]

I think the correct answer is C. 0.2*10^5

8 0

3 years ago

Read 2 more answers

250cm3 of fresh water of density 1000kgm-3 is mixed with 100cm3 of sea water of density 1030kgm-3. Calculate the density of the

natka813 [3]

Answer:

$m_{fresh}= 1000 \frac{Kg}{m^3} * 2.5x10^{-4} m^3 = 0.25 Kg$

And we can do a similar procedure for the sea water:

$m_{sea}= \rho_{sea} V_{sea}$

And after convert the volume to m^3 we got:

$m_{sea}= 1030 \frac{Kg}{m^3} * 1x10^{-4} m^3 = 0.103 Kg$

And then the density for the mixture would be given by:

$\rho_{mixture}= \frac{m_{fresh} +m_{sea}}{v_{fresh} +v_{sea}}$

And replacing we got:

$\rho_{mixt}= \frac{0.25 +0.103 Kg}{2.5x10^{-4} m^3 +1x10^{-4} m^3} = 1008.571 \frac{Kg}{m^3}$

Step-by-step explanation:

For this case we can begin calculating the mass for each type of water:

$m_{fresh}= \rho_{fresh} V_{fresh}$

And after convert the volume to m^3 we got:

$m_{fresh}= 1000 \frac{Kg}{m^3} * 2.5x10^{-4} m^3 = 0.25 Kg$

And we can do a similar procedure for the sea water:

$m_{sea}= \rho_{sea} V_{sea}$

And after convert the volume to m^3 we got:

$m_{sea}= 1030 \frac{Kg}{m^3} * 1x10^{-4} m^3 = 0.103 Kg$

And then the density for the mixture would be given by:

$\rho_{mixture}= \frac{m_{fresh} +m_{sea}}{v_{fresh} +v_{sea}}$

And replacing we got:

$\rho_{mixt}= \frac{0.25 +0.103 Kg}{2.5x10^{-4} m^3 +1x10^{-4} m^3} = 1008.571 \frac{Kg}{m^3}$

6 0

3 years ago

Find the slope of the linear function

il63 [147K]

Answer:

that one over there

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers