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8090 [49]
2 years ago
15

Please help ASAP ITS 14 points

Mathematics
1 answer:
Liono4ka [1.6K]2 years ago
4 0

Answer:

area= 62.88

Step-by-step explanation:

to find the area of the whole figure take the diameter of the circle witch is 8 and multiply it by 11. the area of the whole figure including the semi-circle is 88.

now find the area of the whole circle and divide it in half.

A= pi * r^2 / 2

A= 3.14 * 4^2 /2

A= 3.14 * 16 / 2

A= 25.12

Now take the area of the semi-circle and subtract it from the whole figure.

88-25.12=62.88

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Which of the following rational functions is graphed below?
Cerrena [4.2K]

Answer:

option B

Step-by-step explanation:

We can see in the graph that the function has two values of x where the value of y goes to infinity: x = -6 and x = 6.

These points where the value of the function goes to infinity usually are roots of the polynomial in the denominator of a fraction (when the values of x tend to these values, the denominator of the fraction tends to 0, so we have a discontinuity in the function).

So the option that represents a function that have these points in x = -6 and x = 6 is the function in option B.

The other options show functions that have only one point that goes to infinity.

7 0
3 years ago
In sale the normal price is reduced by 10% Nathalie bought a pair of shoes in the sale of 54 whats the original price
Sholpan [36]
$59.40 was the original price for the shoes

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3 years ago
Does this graph represent a function? Why or why not?
In-s [12.5K]
B. No, because it fails the vertical line test.
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3 years ago
x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

So, the roots are given by:

r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

3 0
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Her total bill was $18.09
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Read 2 more answers
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