A child 4 1/2 feet tall casts a 6-foot shadow. A nearby statue casts a 12-foot shadow.​

Answer:

x = 1

Step-by-step explanation:

given h(x) = - x + 1

when h(x) = 0 , that is

- x + 1 = 0 ( subtract 1 from both sides )

- x + 1 - 1 = 0 - 1 ( simplify both sides )

- x = - 1 ( multiply both sides by - 1 )

- 1 × - x = - 1 × - 1 , that is

x = 1

Answer:

Use the slope-intercept form to find the slope and y-intercept

Any line can be graphed using two points. Select two  x

values, and plug them into the equation to find the corresponding  y

values.  Graph the line using the slope and the y-intercept, or the points.

Slope:  2

y-intercept:  18

N = 5^(10) × 2^(14) × 3

This is about prime factors

By definition, a prime factor is a factor of a number and that factor is also a prime number.

A prime number is one that is divisible by only itself and 1.

The number we have is; 480 × 10^(9)

Now, let's list the prime factors or 480 and they are;

2, 3, 5

Now, using these prime factors alone to get 480, we have;

5 × 3 × 2 × 2 × 2 × 2 × 2 = 480

In powers, gives;

5 × 3 × 2^(5)

Now,the 10^(9) with the 480 can also be expressed in terms of it's prime factors which are 2 and 5 as;

(5 × 2)^(9)

Expanding this gives; 5^(9) × 2^(9) = 10^(9)

Thus;

480 × 10^(9) = 5 × 3 × 2^(5) × 5^(9) × 2^(9)

This gives;

5^(10) × 2^(14) × 3

Read more at; brainly.com/question/4853862

Answer:

I got 5^10 x 2^14 x 3

Step-by-step explanation:

1. Do a tree to find 480 as a product of its prime factors

2. You should get 480= 5x2x3x2x2x2x2

3. 10 expressed as a product of its prime factors is 5x2

4. so 10^9 expressed as a product of its prime factors would be

5x2x5x2x5x2x5x2x5x2x5x2x5x2x5x2x5x2

5. You can then write out 480x10^9 as a product of its prime factors

5x2x3x2x2x2x2x5x2x5x2x5x2x5x2x5x2x5x2x5x2x5x2x5x2

6. change this to use powers

5^10 x 2^14 x 3

Answer:

Sample Response: Approximate between two whole numbers by finding the perfect squares nearest to the target number. Identify which value the non-perfect square root is closest to, then use the iterative process to approximate further to the tenths place, and then further to the hundredths place.