What is equivalent expression of 6x(5+n)

How many zeros are in the product of any nonzero whole number less than 10 and 500

worty [1.4K]

Answer:

If the non-zero whole number is even, then there are 3 zeroes

If the non-zero whole number is odd, then there are 2 zeroes

Step-by-step explanation:

* <em>Lets explain the information in the problem</em>

- Whole numbers are positive numbers, including zero, without

any decimal or fractional

- Nonzero whole numbers are positive numbers, not including zero

- The non-zero whole numbers less then 10 are:

1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9

- The product of any nonzero whole number less than 10 and 500

means multiply any of nonzero number above by 500

- The number of zeros in the product is equal to the sum of the

number of zeros at the end of each of the factors.

# 1 × 500 = 5 × 100 ⇒ 2 zeros

# 2 × 500 = 2 × 5 × 100 = 10 × 100 ⇒ 3 zeros

# 3 × 500 = 3 × 5 × 100 = 15 × 100 ⇒ 2 zeros

# 4 × 500 = 2 × 2 × 5 × 100 = 2 × 10 × 100 ⇒ 3 zeros

# 5 × 500 = 5 × 5 × 100 = 25 × 100 ⇒ 2 zeroes

# 6 × 500 = 2 × 3 × 5 × 100 = 3 × 10 × 100 ⇒ 3 zeros

# 7 × 500 = 7 × 5 × 100 = 35 × 100 ⇒ 2 zeroes

# 8 × 500 = 2 × 2 × 2 × 5 × 100 = 4 × 10 × 100 ⇒ 3 zeros

# 9 × 500 = 3 × 3 × 5 × 100 = 45 × 100 ⇒ 2 zeros

- From all answers above:

# If the non-zero whole number is even (2 , 4 , 6 , 8), then there are

3 zeros in the product

# If the non-zero whole number is odd (1 , 3 , 5 , 7 , 9), then there are

2 zeros in the product

8 0

3 years ago

ASAP GEO PROBLEM HELP NEEDED QUICK

olya-2409 [2.1K]

Answer:

See below.

Step-by-step explanation:

The converse of the Pythagorean Theorem tells us that if the sum of the square of the two shorter lengths is equivalent to the square of the longest length, then the triangle is indeed a right triangle.

We have the side lengths (x²-1), (2x), and (x²+1).

We can see that our hypotenuse will be (x²+1), since it will always yield the largest value for x.

So, we can use the Pythagorean Theorem:

$a^2+b^2=c^2$

Substitute (x²-1) and (2x) for a and b, and substitute (x²+1) for c:

$(x^2-1)^2+(2x)^2\stackrel{?}{=}(x^2+1)^2$

Square each expression:

$(x^4-2x^2+1)+(4x^2)\stackrel{?}{=}(x^4+2x^2+1)$

Combine like terms:

$x^4+(-2x^2+4x^2)+1\stackrel{?}{=}(x^4+2x^2+1)$

Add:

$x^4+2x^2+1\stackrel{\checkmark}{=}x^2+2x^2+1$

We can see that the resulting equations are equivalent, proving that the triangle is indeed a right triangle.

And we're done!

4 0

3 years ago

How do i solve 3x-5y=9;x=-1,0,1

REY [17]

For x = -1, 3(-1) - 5y = 9 => -3 - 5y = 9 => 5y = -3 - 9 = -12 => y = -12/5 = -2.4
for x = 0, 3(0) - 5y = 9 => -5y = 9 => y = 9/-5 = -1.8
for x = 1, 3(1) - 5y = 9 => 3 - 5y = 9 => 5y = 3 - 9 = -6 => y = -6/5 = -1.2

7 0

3 years ago

If f(x) = 3x4 + 2x3 - x + 15,<br> what would be the list of<br> possible rational roots?

Lesechka [4]

Answer: A. ± $\frac{1}{3} ,\frac{5}{3} ,1,3,5,15$

Step-by-step explanation:

± $\frac{1, 3, 5, 15}{1, 3}$

= ± $1, 3, 5, 15, 1/3, 5/3, 5$

Rearranging, we get:

= ± $\frac{1}{3} ,\frac{5}{3} ,1,3,5,15$

7 0

2 years ago

H(x)={(-7,-1),(0,0),(5,8)} domain and range

Bingel [31]

Answer:

The domain is -7, 0, and 5

The range is -1, 0, and 8

Step-by-step explanation:

The domain of a set of points is the x-value. In this case the x-values are -7, 0, and 5 respectively. The range of a set of points is the y-value so in this case the range is -1, 0, and 8.

7 0

3 years ago