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2 years ago

Multiply 2 times 27 using the area model label the model

Mathematics

2 answers:

Viefleur [7K]2 years ago

5 0

Your answer is just 16 I believe. What model are they talking about?

dexar [7]2 years ago

3 0

Sixteennnnnnnnnnnnnn n

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Which problem is solved using this model?

Sonja [21]

A because it start at 20 and divided 1/2 each time. And when you solved 2/ 1/2 , you get 4 and that is correct because you have 4 arrows.

6 0

2 years ago

Solve y'' + 10y' + 25y = 0, y(0) = -2, y'(0) = 11 y(t) = Preview

svetlana [45]

Answer: The required solution is

$y=(-2+t)e^{-5t}.$

Step-by-step explanation: We are given to solve the following differential equation :

$y^{\prime\prime}+10y^\prime+25y=0,~~~~~~~y(0)=-2,~~y^\prime(0)=11~~~~~~~~~~~~~~~~~~~~~~~~(i)$

Let us consider that

$y=e^{mt}$ be an auxiliary solution of equation (i).

Then, we have

$y^prime=me^{mt},~~~~~y^{\prime\prime}=m^2e^{mt}.$

Substituting these values in equation (i), we get

$m^2e^{mt}+10me^{mt}+25e^{mt}=0\\\\\Rightarrow (m^2+10y+25)e^{mt}=0\\\\\Rightarrow m^2+10m+25=0~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{since }e^{mt}\neq0]\\\\\Rightarrow m^2+2\times m\times5+5^2=0\\\\\Rightarrow (m+5)^2=0\\\\\Rightarrow m=-5,-5.$

So, the general solution of the given equation is

$y(t)=(A+Bt)e^{-5t}.$

Differentiating with respect to t, we get

$y^\prime(t)=-5e^{-5t}(A+Bt)+Be^{-5t}.$

According to the given conditions, we have

$y(0)=-2\\\\\Rightarrow A=-2$

and

$y^\prime(0)=11\\\\\Rightarrow -5(A+B\times0)+B=11\\\\\Rightarrow -5A+B=11\\\\\Rightarrow (-5)\times(-2)+B=11\\\\\Rightarrow 10+B=11\\\\\Rightarrow B=11-10\\\\\Rightarrow B=1.$

Thus, the required solution is

$y(t)=(-2+1\times t)e^{-5t}\\\\\Rightarrow y(t)=(-2+t)e^{-5t}.$

6 0

2 years ago

Graph the solution set

antiseptic1488 [7]

Answer:

Graph by first drawing a horizontal line at 3 on the y axis. Anything below this line falls into the y is less than or equal to 3 range. Draw a vertical line at negative 2 on the x axis. This represents x > -2. The next one is a slanted line, so it'll be trickier to graph. If using a graph of 10 - 10, graph from (-8, -10) to (10, 8). If using a graph of 20 - 20, graph from (-18, -20) to (20, 18). As seen in the graph, anything to the upper left of this line is included.

The true inclusion is where all the ranges intersect. Be sure to color this the darkest or with a special pattern.