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2 years ago

Does anybody know the answer to this??

Mathematics

2 answers:

Angelina_Jolie [31]2 years ago

5 0

Answer:

a. 14 Cubic Units

Step-by-step explanation:

This is a complex shape made by a 2x3 and a 2x4 shape

The volume of each of them are 6 cubic units and 8 cubic units respectively.

Together they make a complex shape with the volume of 14 cubic units

lbvjy [14]2 years ago

4 0

Answer:

Step-by-step explanation:

i counted all of the cubes and came up with 14 hence 14 is the answer

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A rectangular swimming pool that is 10 ft wide by 16 ft long is surrounded by a cement sidewalk of uniform width. If the area of

Alchen [17]

We are asked to solve for the width "x" in the given problem. To visualize the problem, see attached drawing.
We have the area of the swimming pool such as:
Area SP = l x w
Area SP = 10 * 16
Area SP = 160 feet2
Area of the swimming pool plus the sidewalk with uniform width:
Area SP + SW = (10 + x) * (16 + x)
160 + 155 = 160 + 10x + 16x + x2
160 -160 + 155 = 26x + x2
155 = 26 x + x2
x2 + 26x -155= 0
Solving for x, we need to use quadratic formula and the answer is 5 feet.

The value of x is <span>5 feet. </span>

5 0

3 years ago

Which of the following is an arithmetic sequence?<br><br>

Lemur [1.5K]

Answer:

either D or A i’m not sure

explanation:

your multiplying by 4 each time.
4 x 4 = 16
16 x 4 = 64
64 x 4 = 256
thats for D

your adding 4 each time
19-15=4
15-11=4
11-7=4
7-3=4
that’s for A

7 0

3 years ago

Please help due in 5 minutes !

Liula [17]

Answer:

all

Step-by-step explanation:

of them

why?

becuase htink

nobody cares

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:

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