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Hunter-Best [27]

3 years ago

5

What is 41/44 times 12/29

1 answer:

Mnenie [13.5K]3 years ago

3 0

The answer is 123/319

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Answer number 4 please need it ASAP

steposvetlana [31]

Prime Factors of 402 =2, 3,67

<span>Prime Factors Tree of 402402
/     \
2     201/       \
3      67</span>

4 0

3 years ago

Read 2 more answers

Identify if the expressions below are polynomials or not for each of them.

monitta

The polynomials can´t contain:

-Variables in the exponents of the terms.
-Variables in the denominator.
- Negative exponents.
-Fractionals exponents.
-The variables cant be as radicals.

Therefore:

a) It is a polynomial (The variable is under a radical)
b) It is not a polynomial.
c) It is a polynomial.
d) It is not a polynomial. (It has a variable in the denominator).
e) It is a polynomial.
f) It is not a polynomial. <span> (It has a </span><span>variable in the denominator).</span>

4 0

3 years ago

Which of these numbers are between 1 and 3/5 <br><br> A. 8/5<br> B. 5/9<br> C. 5/8<br> D. 3/8

monitta

8\5 I think but not sure

5 0

4 years ago

Read 2 more answers

What are the possible solutions to the equation: x-12 = sqrt x+44

Ainat [17]

possible solutions for equation $x-12 = \sqrt{x+44}$ is $x=5 , x=20$

<u>Step-by-step explanation:</u>

Here we have to find the possible solutions to the equation: x-12 = sqrt x+44 .Let's find out:

⇒ $x-12 = \sqrt{x+44}$

⇒ $(x-12)^2 = (\sqrt{x+44})^2$

⇒ $(x^2+144-24x) =x+44$

⇒ $(x^2+144-24x) -(x+44) =0$

⇒ $(x^2+100-25x) =0$

⇒ $x^2-25x+100 =0$

⇒ $x^2-20x-5x+100 =0$

⇒ $x(x-20-5(x-20) =0$

⇒ $(x-5)(x-20)=0$

⇒ $x-5=0 , x-20=0\\$

⇒ $x=5 , x=20$

Therefore , possible solutions for equation $x-12 = \sqrt{x+44}$ is $x=5 , x=20$

6 0

3 years ago

Using the following equation, find the center and radius of the circle. You must show all work and calculations to receive credi

adoni [48]

Answer:

$r = 4$

The center is:

(2, -4)

Step-by-step explanation:

We have the following equation

$x^2 - 4x + y^2 + 8y = -4$

The general equation of a circle has the following form

$(x-h)^2 + (y-k)^2 = r^2$

Where r is the radius and the point (h, k) is the center of the circle

To transform the given equation in the general equation of a circumference we must use the method of square completions

Step 1

$(x^2 - 4x) + (y^2 + 8y) = -4$

<u>Step 2</u>

<u> </u>Add and subtract $(\frac{b}{2})^2$ For an equation of the form

$ax^2 +bx +c$

$(x^2 - 4x +4)-4 + (y^2 + 8y +16) -16= -4$

<u>Step 3 </u> Factor the expressions within the parentheses

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

<u>Step 4 Simplify</u>

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

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

Finally

$r = 4$

The center is:

(2, -4)

7 0

3 years ago