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3 years ago

HELPPP PLSS DUE RN WILL GUVE BRAINLIST

Mathematics

1 answer:

xxMikexx [17]3 years ago

3 0

Answer:

8 × pi

Step-by-step explanation:

The equation for circumference is 2 × pi × r.

Since the diameter is 8, we know the radius is 4.

2 × pi × 4 = 8pi

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Solve for m.<br> 5m = 830<br> m =

mihalych1998 [28]

Answer:

m=166

5m=830

830÷5=166

m=166

4 0

2 years ago

What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.

lapo4ka [179]

Consider the equation $x^6+6x^3+5=0$ .

First, you can use the substitution $t=x^3$ , then $x^6=(x^3)^2=t^2$ and equation becomes $t^2+6t+5=0$ . This equation is quadratic, so

$D=6^2-4\cdot 5\cdot 1=36-20=16=4^2,\ \sqrt{D}=4,\\ \\ t_{1,2}=\dfrac{-6\pm 4}{2} =-5,-1$ .

Then you can factor this equation:

$(t+5)(t+1)=0$ .

Use the made substitution again:

$(x^3+5)(x^3+1)=0$ .

You have in each brackets the expression like $a^3+b^3$ that is equal to $(a+b)(a^2-ab+b^2)$ . Thus,

$x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)$

and the equation is

$(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0$ .

Here $x_1=-\sqrt[3]{5} , x_2=-1$ and you can sheck whether quadratic trinomials have real roots:

1. $D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}$ .

2. $D_2=(-1)^2-4\cdot 1=1-4=-3$ .

This means that quadratic trinomials don't have real roots.

Answer: $x_1=-\sqrt[3]{5} , x_2=-1$

If you need complex roots, then

$x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2} ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}$ .

6 0

3 years ago

0.03 ÷ 10 equals what

gladu [14]

0.03/10 = 0.003 lalalalala its making me type more

5 0

3 years ago

Read 2 more answers

An object is launched from a launching pad 208 ft. above the ground at a velocity of 192ft/sec. what is the maximum height reach

loris [4]

Answer:

The maximum height is 784 feet

Step-by-step explanation:

In this problem we use the kinematic equation of the height h of an object as a function of time

$h(t) = -16t ^ 2 + v_0t + h_0$

Where $v_0$ is the initial velocity and $h_0$ is the initial height.

We know that

$v_0 = 192\ \frac{ft}{sec}$

$h_0 = 208\ ft.$

Then the equation of the height is:

$h(t) = -16t ^ 2 + 192t +208$

For a quadratic function of the form $ax ^ 2 + bx + c$

where $a$

the maximum height of the function is at its vertex.

The vertice is

$x = -\frac{b}{2a}\\\\y = f(\frac{-b}{2a})$

In this case

$a = -16\\b = 192\\c = 208$

Then the vertice is:

$t = -\frac{192}{2(-16)}\\\\t = 6\ sec$

Now we calculate h (6)

$h(6) = -16(6) ^ 2 +192(6) +208\\\\h(6) = 784\ feet$

The maximum height is 784 feet

8 0

3 years ago

Textbook search committee is considering 13 bucks for possible adoption committee has decided to select six of the 13 for furthe

Pani-rosa [81]

Here is a simple way we can do this.
We have six blanks.
__ __ __ __ __ __

Now, we have 13 possible options to fill in blank number one.
13 __ __ __ __ __

Now we have 12 possible options to fill in blank number two because one person has already been chosen.
13 12 __ __ __ __

Now we have 11 possible options for blank number three.
13 12 11 __ __ __

Now we have 10 possible options for blank number four.
13 12 11 10 __ __

And so forth until we get:
13 12 11 10 9 8

Now we just have to multiply the numbers all together.
13 * 12 * 11 * 10 * 9 * 8
is equal to:
1235520 ways.

4 0

3 years ago