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2 years ago

Find the slope on the graph

Mathematics

1 answer:

True [87]2 years ago

4 0

Answer:

im pretty sure it would be 5/3

Step-by-step explanation:

u go up 5 squares and then over 3

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Kelvin can carry a basket 5 feet rachel can carry it 3 feet further than kelvin daniel can carry it half as far s rachel is it r

Ksenya-84 [330]

I think it eould be 17 feet. 5 plus 8 plus 4 is 17.

6 0

3 years ago

When a watch is sold at ksh 126 aloss of x percent is made . If the same ksh 126, a loss of x percent is made . If the same watc

Nitella [24]

Answer:

ksh 140

Step-by-step explanation:

When a watch is sold for ksh 126 a loss of x% is made, that is( 100 - x)%

Hence, the loss amount = (126 X 100)/(100 - x)

When the same watch is sold for ksh 154 a profit of x% is made, that is (100 + x)%

Hence, the profit amout = (154 X 100)/(100 + x)

Therefore, (126 X 100)/(100 - x) = (154 X 100)/(100 + x)

12600(100 + x) = 15400(100 - x)

1260000 + 12600x = 1540000 - 15400x

28000x = 280000

x = 10%

In that case for a watch sold for ksh 126 a loss 10% is made, that is( 100 - 10)%

Hence, loss amount = (126 X 100)/90 = ksh 140

∴ the buying price is ksh 140

7 0

3 years ago

How do you know weather two quantities x and y are proportional?

Colt1911 [192]

Answer:

every difference x-y is the same for related pairs of x and y

8 0

3 years ago

Initially 100 milligrams of a radioactive substance was present. After 6 hours the mass had decreased by 3%. If the rate of deca

Hitman42 [59]

Answer:

The half-life of the radioactive substance is 135.9 hours.

Step-by-step explanation:

The rate of decay is proportional to the amount of the substance present at time t

This means that the amount of the substance can be modeled by the following differential equation:

$\frac{dQ}{dt} = -rt$

Which has the following solution:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.

After 6 hours the mass had decreased by 3%.

This means that $Q(6) = (1-0.03)Q(0) = 0.97Q(0)$ . We use this to find r.

$Q(t) = Q(0)e^{-rt}$

$0.97Q(0) = Q(0)e^{-6r}$

$e^{-6r} = 0.97$

$\ln{e^{-6r}} = \ln{0.97}$

$-6r = \ln{0.97}$

$r = -\frac{\ln{0.97}}{6}$

$r = 0.0051$

$Q(t) = Q(0)e^{-0.0051t}$

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

$Q(t) = Q(0)e^{-0.0051t}$

$0.5Q(0) = Q(0)e^{-0.0051t}$

$e^{-0.0051t} = 0.5$

$\ln{e^{-0.0051t}} = \ln{0.5}$

$-0.0051t = \ln{0.5}$

$t = -\frac{\ln{0.5}}{0.0051}$

$t = 135.9$

The half-life of the radioactive substance is 135.9 hours.

6 0

2 years ago

Hi, I need to send an drawing

lina2011 [118]

Answer:

Ok then send the drawing

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers