What is the feet and in47 in = ? ft ? in

Some one cans help me to find the answer about the whole shape and tell me how do you get the answer step by step please I need

Naya [18.7K]

First let’s call the square shape A and the rectangle shape B.

To find the area of shape A you do 24 x 24 = 576cm squared

To find the area of shape B you do 40 x 24 = 960cm squared

To find the total area you just add the two areas together
576 + 960 = 1536cm squared

4 0

3 years ago

1. Multiply using distributive property.

RoseWind [281]

Answer:

The distributive property tells us how to solve expressions in the form of a(b + c). The distributive property is sometimes called the distributive law of multiplication and division.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

How do you graph 193 + y is greater than or equal to 201

Travka [436]

Answer:

The correct answer is y ≥ 8

Step-by-step explanation:

In order to graph this, we must first solve the equation for y.

193 + y ≥ 201

y ≥ 8

Now that we have the solved version, we can graph on a number line. Draw a closed circle on the point 8 and then draw an arrow to the right to show that it is greater.

8 0

3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$