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3 years ago

¿ Qué porcentaje de 70 es 35?

Mathematics

1 answer:

labwork [276]3 years ago

6 0

Answer:

Step-by-step explanation:

50% mi amigo

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Carlos bought a $235 water heater with his credit card. He used the water heater for five years before replacing it. He paid off

Mazyrski [523]

First we will find the interest on:

P = $235 principal

t = 2 years

r = 0.1415 annual rate

A = future value

I = A - P the interest

A = P(1 + r)^t

A = 235(1 + 0.1415)^2

A = $306.21

I = A - P

I = $306.21 - $235

I = $71.21

the interest was $71.21.

Next lets find the lifetime cost value:

Lifetime cost value = 306.21 + 5*1.56*52 + 5*0.78*52 = $914.61 (considering that 1 year = 52 weeks)

Now lets find the percentage what percentage the interest is of the lifetime cost:

(71.21/914.61)*100 = 7.79%

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3 0

3 years ago

One hour after a hiker left a camp, a cyclist set out to catch up. If the hiker traveled 3 miles per hour and the cyclist travel

katrin [286]

Its 23.1 one because in 23 minutes they travel three miles

7 0

3 years ago

Read 2 more answers

In October sally drove 560 miles in her car.

torisob [31]

The cost is £79.76. Since the car covers 560 miles and 34.5 miles is travelled by one gallon. So, dividing 560 by 34.5 we get 16.23. Since 1 gallon is 4.55 litres, 16.23 gallons is 73.8465. Now the cost of petrol is £1.08 per litre. So, multiplying 73.8465 by £1.08 we have £79.76

4 0

3 years ago

The equation of a curve is y=f(x). the curve goes through the points (1,3) and (3,7). f'(x) = 4x+p, where p is a number.

Nimfa-mama [501]

$f'(x)=4x+p\\
f(x)=2x^2+px+C\\\\
3=2\cdot1^2+p\cdot1+C\\
7=2\cdot3^2+p\cdot3+C\\\\
3=2+p+C\\
7=18+3p+C\\\\
C=1-p\\
3p=-11-C\\\\
3p=-11-(1-p)\\
3p=-11-1+p\\
2p=-12\\
p=-6\\\\
C=1-(-6)\\
C=7\\\\
y=2x^2-6x+7
$

5 0

3 years ago

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

3 years ago