All categories

2 years ago

A kintted pot holder is shaped like a circle. It's radius is 4.5 in. What is the area?

Mathematics

2 answers:

lesya [120]2 years ago

8 0

Answer:

A =63.585 in^2

Step-by-step explanation:

The area of a circle is found by

A = pi r^2

A = pi (4.5)^2

A = pi(20.25)

We can approximate pi by 3.14

A = (3.14) (20.25)

A =63.585 in^2

MakcuM [25]2 years ago

7 0

Answer:

about 63.6 inches

Step-by-step explanation:

formula to find the area of a circle: pi*r^2

pi = 3.14

r = 4.5

3.14*4.5*4.5 = 63.585

You might be interested in

What is the principal if...<br> Rate = 22%<br> Time<br> 1 year<br> Interest = $110

Zanzabum

Answer:

Interest=PTR/100

110 =P×1×22/100

11000=22P

P=500

8 0

2 years ago

I don't know what I'm doing wrong here, can anyone help?

Novosadov [1.4K]

if you look at the part where the first part connects with the second part:

y = 5 if x < - 2

y = -2x + 1 if -2 ≤ x < 1

we don't have a discontinuity there, so there shouldn't be a dot.

<h3></h3><h3>What is wrong with the graph?</h3>

When we graph over intervals like (a, b) or [a, b] or something like that, we use dots to define the end of the intervals, and to denote that the function ends abruptly or we have a jump.

In this case, you can see that between the end and the second part and the beginning of the third part there is a jump, so the use of dots is correct there, but if you look at the part where the first part connects with the second part:

y = 5 if x < - 2

y = -2x + 1 if -2 ≤ x < 1

we don't have a discontinuity there, so there shouldn't be a dot.

That is the only error with the graph.

If you want to learn more about piecewise functions:

brainly.com/question/3628123

#SPJ1

8 0

1 year ago

I really need help asap!!

Firdavs [7]

Given:

A figure of a parallelogram.

The vertex angles are $(12b+8)^\circ,(5b+2)^\circ,2a^\circ$ .

To find:

The values of a and b.

Solution:

We know that the pairs of consecutive angles of a parallelogram are supplementary angles. It mean their sum is 180 degrees.

$(12b+8)^\circ+(5b+2)^\circ=180^\circ$ (Supplementary angles)

$(17b+10)^\circ=180^\circ$

$17b+10=180$

Subtract 10 from both sides.

$17b=180-10$

$17b=170$

Divide both sides by 17.

$b=\dfrac{170}{17}$

$b=10$

Now,

$(5b+2)^\circ=(5(10)+2)^\circ$

$(5b+2)^\circ=(50+2)^\circ$

$(5b+2)^\circ=52^\circ$

And,

$(5b+2)^\circ+2a^\circ=180^\circ$ (Supplementary angles)

$52^\circ+2a^\circ=180^\circ$

$2a^\circ=180^\circ-52^\circ$

$2a^\circ=128^\circ$

Divide both sides by 2.

$a^\circ=\dfrac{128^\circ}{2}$

$a^\circ=64^\circ$

$a=64$

Therefore, the value of a is 64 and the value of b is 10.

5 0

2 years ago

Let S be the solid beneath z = 12xy^2 and above z = 0, over the rectangle [0, 1] × [0, 1]. Find the value of m > 1 so that th

jonny [76]

Answer:

The answer is $\sqrt{\frac{6}{5}}$

Step-by-step explanation:

To calculate the volumen of the solid we solve the next double integral:

$\int\limits^1_0\int\limits^1_0 {12xy^{2} } \, dxdy$

Solving:

$\int\limits^1_0 {12x} \, dx \int\limits^1_0 {y^{2} } \, dy$

$[6x^{2} ]{{1} \atop {0}} \right. * [\frac{y^{3}}{3}]{{1} \atop {0}} \right.$

Replacing the limits:

$6*\frac{1}{3} =2$

The plane y=mx divides this volume in two equal parts. So volume of one part is 1.

Since m > 1, hence mx ≤ y ≤ 1, 0 ≤ x ≤ $\frac{1}{m}$

Solving the double integral with these new limits we have:

$\int\limits^\frac{1}{m} _0\int\limits^{1}_{mx} {12xy^{2} } \, dxdy$

This part is a little bit tricky so let's solve the integral first for dy:

$\int\limits^\frac{1}{m}_0 [{12x \frac{y^{3}}{3}}]{{1} \atop {mx}} \right.\, dx =\int\limits^\frac{1}{m}_0 [{4x y^{3 }]{{1} \atop {mx}} \right.\, dx$

Replacing the limits:

$\int\limits^\frac{1}{m}_0 {4x(1-(mx)^{3} )\, dx =\int\limits^\frac{1}{m}_0 {4x-4x(m^{3} x^{3} )\, dx =\int\limits^\frac{1}{m}_0 ({4x-4m^{3} x^{4}) \, dx$

Solving now for dx:

$[{\frac{4x^{2}}{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right. = [{2x^{2} -\frac{4m^{3} x^{5}}{5} ]{{\frac{1}{m} } \atop {0}} \right.$

Replacing the limits:

$\frac{2}{m^{2} }-\frac{4m^{3}\frac{1}{m^{5}}}{5} =\frac{2}{m^{2} }-\frac{4\frac{1}{m^{2}}}{5} \\ \frac{2}{m^{2} }-\frac{4}{5m^{2} }=\frac{10m^{2}-4m^{2} }{5m^{4}} \\ \frac{6m^{2} }{5m^{4}} =\frac{6}{5m^{2}}$

As I mentioned before, this volume is equal to 1, hence:

$\frac{6}{5m^{2}}=1\\m^{2} =\frac{6}{5} \\m=\sqrt{\frac{6}{5} }$

3 0

2 years ago

the following histogram shows the number of days during the school year that students bought lunch. How many total students are

olga2289 [7]

You have to post a picture of the actually histogram we can’t see it

5 0

2 years ago