All categories

3 years ago

What is the formula for area of a circle? 100 points

Mathematics

2 answers:

wel3 years ago

6 0

Hello!

The formula for the area of a circle is:

A = $\pi$ x r²

sergiy2304 [10]3 years ago

3 0

Area = π r^2 where r = radius of the circle.

You might be interested in

Alexa is planning to drive across the United States in

ivann1987 [24]

Answer:

a=112

b=6

c=280

Step-by-step explanation:

you just plug the number into the equation

28x=y

where x is the gallons

and y is the total miles

8 0

2 years ago

Read 2 more answers

Wrote and solve an equation to find the number.The difference of twice a number and 7 is 9

andrew11 [14]

n - 7 = 9

Step-by-step explanation:

so basically just make up a random variable and place it for the "twice a number" then u sovle it as a 2 step equation

6 0

3 years ago

Which are the solutions of x2 = –7x – 8?

vovangra [49]

Answer:

The solutions are:

$x_1=\frac{-7+\sqrt{17}}{2}$ $x_2=\frac{-7-\sqrt{17}}{2}$

Step-by-step explanation:

We have the following quadratic equation

$x^2 = -7x - 8$

We can rewrite the equation as follows

$x^2+7x + 8=0$

Now we use the quadratic formula to solve the equation

For an equation of the form $ax ^ 2 + bx + c = 0$ the quadratic formula is:

$x=\frac{-b\±\sqrt{b^2-4ac}}{2a}$

In this case:

$a=1,\ b=7,\ c=8$

Then:

$x=\frac{-7\±\sqrt{7^2-4(1)(8)}}{2(1)}$

$x=\frac{-7\±\sqrt{49-32}}{2}$

$x=\frac{-7\±\sqrt{17}}{2}$

$x_1=\frac{-7+\sqrt{17}}{2}$

$x_2=\frac{-7-\sqrt{17}}{2}$

7 0

3 years ago

Read 2 more answers

The polynomial x^2+3x-1 is a factor of x^4+3x^3-2x^2-3x+1 true or false?

pashok25 [27]

Answer:

Yes, it is true that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ .

Step-by-step explanation:

Let us try to factorize $x^4+3x^3-2x^2-3x+1$

$x^4+3x^3-2x^2-3x+1\\\Rightarrow x^4-2x^2+1-3x+3x^3$

Let us try to make a whole square of the given terms:

$\Rightarrow (x^2)^2-2\times x^2 \times 1+1^2+3x^3-3x\\\Rightarrow (x^2-1)^2+3x^3-3x\\$

--------------

Formula used above:

$a^{2} -2 \times a \times b +b^{2} = (a-b)^2$

In the above equation, we had $a = x, b = 1$ .

--------------

Further solving the above equation, taking $3x$ common out of $3x^3-3x$

$\Rightarrow (x^2-1)^2+3x(x^2-1)\\$

Taking $(x^{2} -1)$ common out of the above term:

$\Rightarrow (x^2-1)((x^2-1)+3x)\\\Rightarrow (x^2-1)(x^2+3x-1)$

So, the two factors are $(x^2-1)\ and\ (x^2+3x-1)$ .

$\therefore$ The statement that $x^2+3x-1$ is a factor of $x^4+3x^3-2x^2-3x+1$ is <em>True.</em>

7 0

3 years ago

Complete the explanation on the 2 errors Margaret made, and then give the correct answer.

klemol [59]

Answer:

For the first page 2nd line they did 88846 squared instead of cubed.

The correct answer is 935,087,000,000,000.

Step-by-step explanation:

8 0

2 years ago