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3 years ago

What name describe a triangle

Mathematics

2 answers:

Goryan [66]3 years ago

5 0

Can be described as a "Shape"

vodomira [7]3 years ago

3 0

Acute, Obtuse, Isosceles, and Right triangles are all shapes.

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Initially, there were only 197 weeds at a park. The weeds grew at a rate of 25% each week. The following function represents the

olganol [36]

Answer:

none of the above

f(x) ≈ 197·1.03^x; approximately 3% daily

Step-by-step explanation:

If we let x represent days, then x/7 represents weeks and we can rewrite f(x) as ...

f(x/7) = 197·1.25^(x/7) = 197·(1.25^(1/7))^x

f(x) ≈ 197·1.03^x

The daily multiplier of 1.03 represents a daily growth rate of

1.03 -1 = .03 = 3%

_____

These answers are not found among the offered choices:

f(x) = 197·1.03^x

3% daily growth

Step-by-step explanation:

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3 years ago

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Ana's manager is trying to figure out how much to charge for a chair that just arrived. If the wholesale price of the chair is $

Veronika [31]

Answer: 249.75

Step-by-step explanation: 85% of 135 =114.75 +135=249.75

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3 years ago

Whatttttt? Help meeeeeeeeeeeeeeeeeeeeeeeee... lost all braincells binge watching peppa!!!

morpeh [17]

Answer: C

Step-by-step explanation: It's the most logical.

and don't lie. We all know you watch dora more.

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Can yall check my answer ?

Rzqust [24]

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Step-by-step explanation:

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) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago