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3 years ago

Configurative Area Problems Find the area of the shaded region. BRAINLIEST

Mathematics

2 answers:

natali 33 [55]3 years ago

8 0

Answer:

78.54 cm

formula: A= pi x r^2

babunello [35]3 years ago

7 0

Area of 1/4 = 1/4 x 3.14 x 10 x 10 = 78.5cm2
area of 2 semi circles = 3.14x5x5= 78.5 cm2
78.5+78.5=157cm2

formula: π x r x r = area of circle
1/4 x π x r x r = area of 1/4 of a circle
x = times

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Select the correct answer from each drop-down A cable on a suspension bridge can be modeled by this equation, where h is the cab

FromTheMoon [43]

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3 years ago

21. The equation x² + y² - 6x + 8y = 74 represents circle C. What is the radius of circle C?

marin [14]

Answer:

The radius is $3\sqrt{11}$ .

Step-by-step explanation:

The standard equation for a circle is $(x-h)^2+(y-k)^2=r^2$ .

I like to call this form center-radius form because it tells you the center (h,k) and the radius r.

So we put it in this form to determine r, the radius, which is the desired value here.

We will need to complete the square.

Thankfully there is a formula to take more thinking out of the process:

$x^2+dx+(\frac{d}{2})^2=(x+\frac{d}{2})^2$ .

Let's begin:

$x^2+y^2-6x+8y=74$

Reorganize the equation so the x's are together and the y's are together.

$x^2-6x+y^2+8y=74$

Now time to being completing the square. Keep in mind whatever you add to one side you need to add the other side to keep the equation equivalent to what it once was.

$x^2-6x+(\frac{-6}{2})^2+y^2+8y+(\frac{8}{2})^2=74+(\frac{-6}{2})^2+(\frac{8}{2})^2$

Use the other side of the formula I mentioned for completing the square:

$(x+\frac{-6}{2})^2+(y+\frac{8}{2})^2=74+(-3)^2+(4)^2$

$(x-3)^2+(y+4)^2=74+9+16$

$(x-3)^2+(y+4)^2=74+25$

$(x-3)^2+(y+4)^2=99$

So the center is (3,-4).

The radius is $\sqrt{99}$ .

You can actually simplify the radius since 99 contains a factor that is a perfect square.

$\sqrt{99}=\sqrt{9}\sqrt{11}=3\sqrt{11}$

7 0

4 years ago

Read 2 more answers

Help me find the difference!! ☹️

Marysya12 [62]

The differences are that in the first equation it has an x and the other has a 3. The first equation is divided by x+3 while the other is divided by x-2