I have 7 hundreds blocks, 5 tens blocks, and 8 ones blocks. I use my blocks to model two 3-digit numbers. What could my two numbers be?

I just had the same question on a test and the answer was 90 MPH. Hope this helps somebody!

600 can be written as [tex]\(2^3 \times 3^1 \times 5^2\)[/tex], where 2, 3, and 5 are prime numbers.

To express 600 as the product of prime numbers, we'll use prime factorization.

Prime factorization involves breaking down a number into its prime factors.

Here's how we can do it:

**Start with the smallest prime number, 2:**

  [tex]\( 600 \div 2 = 300 \)[/tex]

  [tex]\( 300 \div 2 = 150 \)[/tex]

  [tex]\( 150 \div 2 = 75 \)[/tex]

**Next, continue with the next smallest prime number, 3:**

  [tex]\( 75 \div 3 = 25 \)[/tex]

  [tex]\( 25 \div 5 = 5 \)[/tex]

**Now, we can't divide further by smaller prime numbers, so we try dividing by the next smallest prime, 5:**

  [tex]\( 5 \)[/tex] is already a prime number.

**There are no more prime factors to consider, so we stop.**

Now, let's write down the prime factors we obtained:

[tex]\[ 600 = 2 \times 2 \times 2 \times 3 \times 5 \times 5 \][/tex]

So, [tex]\( a = 2 \), \( b = 2 \), \( c = 3 \), and \( d = 5 \).[/tex]

Therefore, we can write:

[tex]\[ 600 = 2 \times 2 \times 2 \times 3 \times 5 \times 5 \][/tex]

Thus, [tex]\( 600 \)[/tex] can be written as [tex]\( 2^3 \times 3^1 \times 5^2 \)[/tex], where [tex]\( 2, 3, \) and \( 5 \)[/tex] are all prime numbers.

597 is a greater value than 579.

A number pattern is a pattern in a series of numbers that represents the common relationship between the numbers.

Rounding some number to a specific value is making its value simpler mostly done for better readability or accessibility.

Rounding to some place keeps it accurate on the left side of that place but rounded or sort of like trimmed from the right in terms of exact digits.

We have to find How are the numbers 579 and 597 different to each other.

As we can see that the last two digit of the numbers 79 and 97 are flipped.

We know that 97 is greater than 79.

Therefore, 597 is a greater value than 579.

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A because you are multiplying 3/4 by 1/2, you make the product smaller than 3/4

B beacue any number multiplied by one remains that number, since you are multiplying 3/4 by 1 it will remaim 3/4

C because the number multiplied by 3/4 is larger than 1 ( 3/2 = 1 1/2) the product will be larger than 3/4

Function notation and ordered pair solutions of a function are closely connected. Function notation is a way to represent a function by its name and its input value, while an ordered pair solution of a function is a pair of numbers (x, y) such that y is the output of the function when x is the input.

To evaluate a function using function notation, we simply substitute the input value into the function's expression. For example, if the function is f(x) = x^2, then to evaluate f(2), we would substitute 2 into the function's expression:

f(2) = 2^2 = 4

This means that the ordered pair solution (2, 4) is a solution of the function f(x) = x^2.

In general, any ordered pair solution of a function can be evaluated using function notation. For example, if the function is g(x) = 2x + 1, and the ordered pair solution is (3, 7), then we can evaluate g(3) as follows:

g(3) = 2(3) + 1 = 6 + 1 = 7

This confirms that the ordered pair solution (3, 7) is a solution of the function g(x) = 2x + 1.

Conversely, any function value can be represented as an ordered pair solution of the function. For example, if the function is h(x) = x^3, and the function value is 8, then we can represent this as the ordered pair solution (2, 8), since h(2) = 8.

In general, any function value can be represented as an ordered pair solution of the function by writing the input value as the first coordinate and the function value as the second coordinate.

Therefore, function notation and ordered pair solutions of a function are closely connected. Function notation is a way to represent a function and its input value, while an ordered pair solution of a function is a pair of numbers (x, y) such that y is the output of the function when x is the input. Any function value can be evaluated using function notation, and any ordered pair solution of a function can be represented as a function value.

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The following question may be like this:

What is the connection between function notation to evaluate a function at certain values and ordered pair solutions of the​ function?

The number of pounds of coffee used is 30 pounds.

Given data:

Let x be the number of pounds of type A coffee used.

Since the blend uses three times as many pounds of type B coffee as type A, the number of pounds of type B coffee used is 3x.

The cost of type A coffee is $4.50 per pound, so the cost of x pounds of type A coffee is 4.50x dollars.

The cost of type B coffee is $5.50 per pound, so the cost of 3x pounds of type B coffee is 5.50 * 3x = 16.50x dollars.

The total cost of the blend is $634.50, so the equation is:

4.50x + 16.50x = 634.50

Combine like terms:

21x = 634.50

On solving for x:

x = 634.50 / 21

x = 30

Hence, 30 pounds of type A coffee were used in the blend.

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Answer:

[tex]3ax+3b^2-3c+2[/tex]

Step-by-step explanation:

We have been given an expression [tex]3(ax+b^2-c)+2[/tex]. We are asked to remove the parenthesis and combine like terms for our given expression.

Using distributive property [tex]a(b+c)=ab+ac[/tex], we will get:

[tex]3*ax+3*b^2-3*c+2[/tex]

[tex]3ax+3b^2-3c+2[/tex]

We can see that our given expression don't have like terms, therefore, our expression would be [tex]3ax+3b^2-3c+2[/tex].

Answer:

Option A : [tex]\frac{log(x+2)}{log(4)}[/tex]

Step-by-step explanation:

Write the expression when the change of base formula is applied to log4(x+2

Given [tex]log_4(x+2)[/tex]

USe change of base formula

[tex]log_b(a)= \frac{log(a)}{log(b)}[/tex]

WE apply change of base formula for the given log

the base of log becomes the denominator .

Numerator becomes the x+2

[tex]log_4(x+2)= \frac{log(x+2)}{log(4)}[/tex]

option A is the correct answer

Final answer:

The function g(x), being different for rational and irrational numbers, is nowhere continuous on the set of real numbers because it cannot satisfy the condition for continuity at any point.

Explanation:

We are asked to determine for what values of x the function g(x) is continuous. The function is defined to be 0 if x is rational, and 4x if x is irrational, over the set of real numbers. First, let's understand a key concept: for a function to be continuous at a point, the limit of the function as it approaches the point from either direction must be equal to the function's value at that point. In the case of g(x), the function takes the value of 0 for all rational numbers, but instantly changes to 4x for irrational numbers which are densely populated around every rational number.

Since the value of g(x) for rationals (0) is different from the nearby values for irrationals (which would be non-zero and depend on x), g(x) does not have a limit that equals its value at rational points and thus cannot be continuous at any rational number. Moreover, at irrational values of x, we cannot have continuity either, as any neighborhood around an irrational number contains rational numbers where g(x) would suddenly jump to 0, disrupting the limit process once again.

Therefore, the function g(x) is nowhere continuous on the set of real numbers since it cannot satisfy the condition for continuity at any point. Hence, the correct answer to the given question is 'none of these'.

Answer:

Option B is correct

x= 3, At level 3, 27  number of people contacted

Step-by-step explanation:

As per the given statement:

The number of people contacted at each level of a phone tree can be represented by:

[tex]f(x) = 3^x[/tex]

where, x represents the level.

We have to find the value of x when f(x) = 27.

⇒27 = 3^x

We can write 27 as:

[tex]27 = 3 \times 3 \times 3 = 3^3[/tex]

then;

[tex]3^3 = 3^x[/tex]

on comparing we have;

3 = x

or

x = 3

Therefore, the value of x is, 3.

The area of the region bounded by the parabola[tex]y=2x^2[/tex]angent line at (5,50), and the x-axis is found by integrating the function of the parabola, finding the x-intercept of the tangent line, and subtracting the area under the tangent line from the area under the parabola between x=0 and x=5.

The area of the region bounded by the parabola , the tangent line at the point (5,50), and the x-axis can be found using integral calculus. First, we find the equation of the tangent line to the parabola at (5,50). The derivative of y with respect to x is given by [tex]y=2x^2[/tex]= 4x. At x=5, the slope of the tangent line is 20. Thus, the equation of the tangent line is y - 50 = 20(x - 5). To find the bounded area, we integrate the area under the parabola from the x-intercept of the tangent line to x=5, and subtract the area under the tangent line in the same interval.

Let's call the x-intercept of the tangent line x1. To find x1, we set the y-value of the tangent line equation to 0 and solve for x. The integration itself uses the antiderivative of  which is [tex]2x^2,[/tex] and the antiderivative of the linear tangent line equation. Subtracting the integral of the tangent line from the integral of the parabola gives us the exact area under the parabola.

(450-210)/4+50=110

 hope this helps