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3 years ago

13

What is the circumference of this circle?

1 answer:

Yanka [14]3 years ago

7 0

C ≈ 307.88
Hope that helps.

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14 = 2y/5 what is the solution?

Gnom [1K]

Answer:

y=35

Step-by-step explanation:

y in (-oo:+oo)

14 = (2*y)/5 // - (2*y)/5

14-((2*y)/5) = 0

(-2/5)*y+14 = 0

14-2/5*y = 0 // - 14

-2/5*y = -14 // : -2/5

y = -14/(-2/5)

y = 35

y = 35

5 0

3 years ago

What is the radius of a circle with an area of 50.24 cubic inches? Use 3.14 for pi. Enter your answer in the box.

miss Akunina [59]

Answer:

r=4

Step-by-step explanation:

Area of a circle is

A = pi* r^2

We know the area and 3.14 = pi

50.24 = 3.14 * r^2

Divide each side by 3.14

50.24/3.14 =3.14 /3.14 *r^2

16 = r^2

Take the square root of each side

sqrt(16) = sqrt(r^2)

4 =r

3 0

3 years ago

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the<br> function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

$f(x)=(x-2)^{2}+1$

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

<u>Then, the minimum point of this function is (3,1) and it corresponds to (A)</u>

46)

Let's use the same method here.

$g(x)=f(x+2)$

$g(x)=((x+2)-2)^{2}+1$

$g(x)=(x)^{2}+1$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

<u>Then, the minimum point of this function is (0,1) and it corresponds to (C)</u>

47)

Let's use the same method here.

$g(x)=f(x)+2$

$g(x)=(x-2)^{2}+1+2$

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

<u>Then, the minimum point of this function is (2,3) and it corresponds to (B)</u>

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

$g(x)=(x-2)^{2}-2$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}-2$

$g(2)=-2$

<u>Then, the minimum point of this function is (2,-2) and it corresponds to (D)</u>

<u />

I hope it helps you!

<u />

8 0

3 years ago

Area of the figure??

8_murik_8 [283]

Answer:

28.5 sq. meters

Step-by-step explanation:

6x3=18 + 6 x 2 divided by 2=6 + 3x3 divided by 2=4.5

=28.5

8 0

3 years ago

Edg vector operations, any help appreciated!

viktelen [127]

$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \: Add \:\: -6 \hat i - 6\hat j \:\:with \:\; Vector \:\; c$

____________________________________

$\large \tt Solution \: :$

Vector d can be represented as :

$\qquad \tt \rightarrow \: - 2 \hat i - 2 \hat j$

Vector c can be represented as :

$\qquad \tt \rightarrow \: 4 \hat i + 4\hat j$

we have to create vector d from vector c

So, let's assume a vector x, such that sum of vector x and vector c equals to vector d

$\qquad \tt \rightarrow \: x + ( 4 \hat i + 4 \hat j) = - 2 \hat i - 2 \hat j$

$\qquad \tt \rightarrow \: x = - ( 4 \hat i + 4 \hat j) - 2 \hat i - 2 \hat j$

$\qquad \tt \rightarrow \: x = (- 4 \hat i - 2 \hat i) + ( - 4 \hat j - 2 \hat j)$

$\qquad \tt \rightarrow \: x = - 6 \hat i -6 \hat j$

Henceforth, in order to get vector d, we need to add (-6i - 6j) in vector c

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

8 0

2 years ago