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1 year ago

5

Transformation example

1 answer:

leonid [27]1 year ago

4 0

Answer:

Rotation, Reflection, Dilation.

Step-by-step explanation:

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Determine if the given side lengths could be used to form a unique triangle, many different triangles, or no triangles. explain

ehidna [41]

Given sides of length 300 mm or 190 mm, you can form 4 different triangles:

equilateral with sides 190 mm
equilateral with sides 300 mm
isosceles with two sides 190 mm and one side 300 mm
isosceles with two sides 300 mm and one side 190 mm

The triangle inequality requires the sum of the lengths of any two sides be not less than the length of the third side. Since 2×190 > 300, you can mix and match these side lengths any way you want. With three sides and 2 choices for each, there are only a limited number of possibilities.

In the above, we have not listed ones that are simply rotations or reflections of a congruent triangle. (A 190-190-300 triangle looks the same as a 190-300-190 triangle, for example.)

6 0

2 years ago

The use of mathematical methods to study the spread of contagious diseases goes back at least to some work by Daniel Bernoulli i

harina [27]

Answer:

a

$y(t) = y_o e^{\beta t}$

b

$x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

c

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

Step-by-step explanation:

From the question we are told that

$\frac{dy}{y} = -\beta dt$

Now integrating both sides

$ln y = \beta t + c$

Now taking the exponent of both sides

$y(t) = e^{\beta t + c}$

=> $y(t) = e^{\beta t} e^c$

Let $e^c = C$

So

$y(t) = C e^{\beta t}$

Now from the question we are told that

$y(0) = y_o$

Hence

$y(0) = y_o = Ce^{\beta * 0}$

=> $y_o = C$

So

$y(t) = y_o e^{\beta t}$

From the question we are told that

$\frac{dx}{dt} = -\alpha xy$

substituting for y

$\frac{dx}{dt} = - \alpha x(y_o e^{-\beta t })$

=> $\frac{dx}{x} = -\alpha y_oe^{-\beta t} dt$

Now integrating both sides

$lnx = \alpha \frac{y_o}{\beta } e^{-\beta t} + c$

Now taking the exponent of both sides

$x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} + c}$

=> $x(t) = e^{\alpha \frac{y_o}{\beta } e^{-\beta t} } e^c$

Let $e^c = A$

=> $x(t) =K e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

Now from the question we are told that

$x(0) = x_o$

So

$x(0)=x_o =K e^{\alpha \frac{y_o}{\beta } e^{-\beta * 0} }$

=> $x_o = K e^{\frac {\alpha y_o }{\beta } }$

divide both side by $(K * x_o)$

=> $K = x_o e^{\frac {\alpha y_o }{\beta } }$

So

$x(t) =x_o e^{\frac {-\alpha y_o }{\beta } } * e^{\alpha \frac{y_o}{\beta } e^{-\beta t} }$

=> $x(t)= x_o e^{\frac{-\alpha * y_o }{\beta} + \frac{\alpha y_o}{\beta } e^{-\beta t} }$

=> $x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }$

Generally as t tends to infinity , $e^{- \beta t}$ tends to zero

so

$\lim_{t \to \infty} x(t) = x_oe^{\frac{-\alpha y_o}{\beta } }$

5 0

3 years ago

The local kennel has 3 spots for cats to every 5 dogs. if the local kennel has a total of 72 spots for both cats and dogs, then

aliina [53]

3:5
5+3= 8
72 / 8 = 9
9 x 3 = 27
27 spots for cats

5 0

3 years ago

Find the least common denominator for these two rational expressions. B/b^2 -64 -7b/b^2+7b-8

san4es73 [151]

Answer:

The least common denominator is (b-8)(b+8)(b-1)

Step-by-step explanation:

We are given expression as

$\frac{b}{b^2-64}-\frac{7b}{b^2+7b-8}$

Firstly, we will factor both denominators

$b^2-64=b^2-8^2=(b-8)(b+8)$

$b^2+7b-8=(b+8)(b-1)$

so, we can plug it back

$\frac{b}{(b-8)(b+8)}-\frac{7b}{(b+8)(b-1)}$

First term denominator is

(b-8)(b+8)

Second term denominator is

(b+8)(b-1)

So,

Least common denominator will be

(b-8)(b+8)(b-1)

So, we get

$LCD=(x-8)(x+8)(x-1)$

3 0

2 years ago

Round to the nearest tenth 6.2x+1.2=8.9

Anuta_ua [19.1K]

Answer:

x = 1.2...

Step-by-step explanation:

Start with the given equation and isolate the variable.

6.2x + 1.2 = 8.9

Subtract 1.2 from each side:

6.2x = 7.7

Divide each side by 6.2:

x = 1.2...

7 0

3 years ago