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andrezito [222]
3 years ago
7

Hehe need tons of help today

Mathematics
1 answer:
Rasek [7]3 years ago
4 0

A plot of the points shows the polygon is a square with side length 3 units. The area is ...

... 3 units × 3 units = 9 square units

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ANSWER PLEASE QUICK!! I AM ON A TIMERRR
Fynjy0 [20]

Answer:

smaller x is  to the power of 2 3x is the larger.

Step-by-step explanation:

5 0
3 years ago
Select the expression that is equivalent to the polynomial given below.<br> (4x+3)(3x-8)
posledela

Answer:

Step-by-step explanation:

5 0
2 years ago
I will mark brainliest!!!!
nata0808 [166]

Answer:

24 ft^2

Step-by-step explanation:

The small aquarium:

1 * 3 * 2 = 6 ft^2

The large aquarium

(1 + 1) * (3 + 2) * (2 + 1)

2 * 5 * 3 = 30 ft^2

Difference between the larger aquarium and the smaller aquarium:

30 - 6, or 24.

This means that the answer to this question is a, 24 ft^2

5 0
2 years ago
A circle has a radius of 8 cm. What is its area?<br><br> Use <img src="https://tex.z-dn.net/?f=%20%5Cpi%20" id="TexFormula1" tit
jekas [21]
The formula for the area of the circle is:
Area=  \pi r^{2}
All you have to do is plug in 8 for r :)
7 0
3 years ago
Solve y" + y = tet, y(0) = 0, y'(0) = 0 using Laplace transforms.
irina1246 [14]

Answer:

The solution of the diferential equation is:

y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}

Step-by-step explanation:

Given y" + y = te^{t}; y(0) = 0 ; y'(0) = 0

We need to use the Laplace transform to solve it.

ℒ[y" + y]=ℒ[te^{t}]

ℒ[y"]+ℒ[y]=ℒ[te^{t}]

By using the Table of Laplace Transform we get:

ℒ[y"]=s²·ℒ[y]+s·y(0)-y'(0)=s²·Y(s)

ℒ[y]=Y(s)

ℒ[te^{t}]=\frac{1}{(s-1)^{2}}

So, the transformation is equal to:

s²·Y(s)+Y(s)=\frac{1}{(s-1)^{2}}

(s²+1)·Y(s)=\frac{1}{(s-1)^{2}}

Y(s)=\frac{1}{(s^{2}+1)(s-1)^{2}}

To be able to separate in terms, we use the partial fraction method:

\frac{1}{(s^{2}+1)(s-1)^{2}}=\frac{As+B}{s^{2}+1} +\frac{C}{s-1}+\frac{D}{(s-1)^2}

1=(As+B)(s-1)² + C(s-1)(s²+1)+ D(s²+1)

The equation is reduced to:

1=s³(A+C)+s²(B-2A-C+D)+s(A-2B+C)+(B+D-C)

With the previous equation we can make an equation system of 4 variables.

The system is given by:

A+C=0

B-2A-C+D=0

A-2B+C=0

B+D-C=1

The solution of the system is:

A=1/2 ; B=0 ; C=-1/2 ; D=1/2

Therefore, Y(s) is equal to:

Y(s)=\frac{s}{2(s^{2} +1)} -\frac{1}{2(s-1)} +\frac{1}{2(s-1)^{2}}

By using the inverse of the Laplace transform:

ℒ⁻¹[Y(s)]=ℒ⁻¹[\frac{s}{2(s^{2} +1)}]-ℒ⁻¹[\frac{1}{2(s-1)}]+ℒ⁻¹[\frac{1}{2(s-1)^{2}}]

y(t)=\frac{1}{2}cos(t)- \frac{1}{2}e^{t}+\frac{t}{2} e^{t}

8 0
3 years ago
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