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3 years ago

A student earns $9 for 1 h and $36 for 4 h

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

7 0

I do not the question. Please explain

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On the world map,the scale is 1/5= 636 miles.The distance is 2 and 1/8 inches.How many miles is it?

crimeas [40]

The answer for this problem is that there is 6757.5 miles for 2 and 1/8 inches

6 0

3 years ago

A population of insects increases at the rate of 200 10t 13t2 what is the change in the population of insects between day 0 and

RUDIKE [14]

Answer:

762 days

Step-by-step explanation:

Given

$Rate = 200 + 10t + 13t^2$

Let the rate be R.

So the rate change with time is represented as:

$\frac{dR}{dt} = 200 +10t + 13t^2$

So:

$dR = (200 + 10t + 13t^2)\ dt$

To get the number of insects between day 0 and day 3, we need to integrate dR and set the bounds to 0 and 3

i.e.

$dR = (200 + 10t + 13t^2)\ dt$ becomes

$\int\limits^3_0 {dR} \, =\int\limits^3_0 (200 + 10t + 13t^2)\ dt$

$R =\int\limits^3_0 (200 + 10t + 13t^2)\ dt$

Integrate

$R = 200t + \frac{10t^2}{2} + \frac{13t^3}{3} [3,0]$

Solve for R by substituting 0 and 3 for t

$R = (200*3 + \frac{10*3^2}{2} + \frac{13*3^3}{3}) - ( 200*0 + \frac{10*0^2}{2} + \frac{13*0^3}{3})$

$R = (200*3 + \frac{10*9}{2} + \frac{13*27}{3}) - (0 + \frac{10}{2} + \frac{0}{3})$

$R = (200*3 + 5*9 + 13*9) - (0)$

$R = 600 + 45 + 117 - 0$

$R =762$

The population of insect between the required interval is 762

4 0

3 years ago

CKPenny is selling 50 coats. Each coat with a hood have 5 buttons and each coat without a hood have 10 buttons. If there are 400

dexar [7]

Answer:

40 coats don't have hoods

7 0

3 years ago

If a +2<b+3 then ____. a-b<1 a<b+1 a<b b<a

OLga [1]

Answer:

a<b+1

Step-by-step explanation:

a +2<b+3

a<b+3-2

a<b+1

6 0

2 years ago

Read 2 more answers

One of the legs of a right triangle measures 12 cm and the other leg measures 9 cm.

PolarNik [594]

Answer:

15cm

Step-by-step explanation:

a^2+b^2=c^2

12^2+9^2=c^2

144+81=c^2

c^2=225

c=15

7 0

2 years ago

Read 2 more answers