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2 years ago

The radius of a circle is 7 cm.the circle is divided into two equal parts.what is the perimeter of each semicircle part

Mathematics

1 answer:

pantera1 [17]2 years ago

8 0

Answer:

≈ 22 cm

Step-by-step explanation:

the circumference (C) of a circle ( the perimeter ) is calculated as

C = 2πr ( r is the radius )

= 2π × 7 = 14π

since divided into 2 equal parts then

$\frac{1}{2}$ C = 14π ÷ 2 = 7π ≈ 22 cm ( to the nearest whole number )

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Enter the value of a and b that complete the sum: 3/x + 5/x^2 = ax+b/x^2

Aleksandr [31]

Answer:

a= $\frac{3}{x^{2} }$

b=5

Step-by-step explanation:

we are given

$\frac{3}{x} }$ + $\frac{5}{x^{2} }$ =ax+ $\frac{b}{x^{2} }$

we will compare both sides of equaion

$\frac{b}{x^{2} }$ = $\frac{5}{x^{2} }$

comparing both sides we get

b=5

and

$\frac{3}{x} }$ =ax

a= $\frac{3}{x^{2} }$

hence

a= $\frac{3}{x^{2} }$

and b=5

4 0

4 years ago

Find the function y1 of t which is the solution of 121y′′+110y′−24y=0 with initial conditions y1(0)=1,y′1(0)=0. y1= Note: y1 is

strojnjashka [21]

Answer:

Step-by-step explanation:

The original equation is $121y''+110y'-24y=0$ . We propose that the solution of this equations is of the form $y = Ae^{rt}$ . Then, by replacing the derivatives we get the following

$121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)$

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

$121r^2+110r-24=0$

Recall that the roots of a polynomial of the form $ax^2+bx+c$ are given by the formula

$x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}$

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

$r_1 = -\frac{12}{11}$

$r_2 = \frac{2}{11}$

So, in this case, the general solution is $y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}$

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

$c_1 + c_2 = 1$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 0$ (or equivalently $c_2 = 6c_1$

By replacing the second equation in the first one, we get $7c_1 = 1$ which implies that $c_1 = \frac{1}{7}, c_2 = \frac{6}{7}$ .

So $y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}$

b) By using y(0) =0 and y'(0)=1 we get the equations

$c_1+c_2 =0$

$c_1\frac{-12}{11} + c_2\frac{2}{11} = 1$ (or equivalently $-12c_1+2c_2 = 11$

By solving this system, the solution is $c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}$

Then $y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}$

The Wronskian of the solutions is calculated as the determinant of the following matrix

$\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2$

By plugging the values of $y_1$ and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

$e^{\int -p(x) dx}$

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

$e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}$

Note that this function is always positive, and thus, never zero. So $y_1, y_2$ is a fundamental set of solutions.

8 0

3 years ago

Ken’s predicted temperature for April is 7ºC. His prediction for May is 2ºC higher than April. He used the number line to solve

Gekata [30.6K]

Answer:

He should have moved to the right.

Step-by-step explanation:

Ken moved 7 to the right which is correct because he needs to get 7º higher, but when he moved 2º to the left that would be a temperature decrease which is is incorrect.

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3 years ago

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bulgar [2K]

The answer would be b

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3 years ago

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3241004551 [841]

Answer:

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