Samuel recorded the number of different types of pets his friends have in the table shown below: Cats 5 Dogs 3 Rabbits 1 Guinea Pigs 3

Answer:

(m-n)(r-s)

Step-by-step explanation:

We will factor this by grouping.  First we rearrange the terms so the m's are together and the n's are together:

mr-ms+ns-nr

Next we group the first two and the last two:

(mr-ms)+(ns-nr)

The GCF of the first group is m; factor this out:

m(r-s)+(ns-nr)

The GCF of the second group is n; factor this out:

m(r-s)+n(s-r)

We want to have the same thing in both sets of parentheses; to do this, we will multiply the second factor by -1:

m(r-s)+-n(r-s)

Factoring out the term now in common, r-s, we have

(r-s)(m-n) = (m-n)(r-s)

Answer:

PANCKACKES

Step-by-step explanation:

2x-20 = x+8

x-20 =8

x = 28

28+8 =36

 angle B = 36 degrees

answer is b

Answer:

A - 6$

You pay six dollars up front, on day zero.

By setting up a system of equations and solving for the number of bicycles (B) and tricycles (T) based on the total seats and wheels, we can determine there are 27 bicycles and 5 tricycles.

To determine the number of bicycles and tricycles in the bicycle store given that there are 32 seats and 69 wheels, we can set up a system of equations. Let's let B represent the number of bicycles and T represent the number of tricycles. Since bicycles have 2 wheels and tricycles have 3 wheels, we can express the total number of wheels as:

2B + 3T = 69

Each bicycle and tricycle has one seat, so we can also express the total number of seats as:

B + T = 32

To solve this system, we can multiply the second equation by 2 and subtract it from the first equation to eliminate B:

(2B + 3T) - (2(B + T)) = 69 - 64

2B + 3T - 2B - 2T = 69 - 64

T = 5

Now we substitute T into the second equation:

B + 5 = 32

B = 32 - 5

B = 27

So, there are 27 bicycles and 5 tricycles in the bicycle store.

Part 1 (or a if you prefer):

f(x) = 5^x

f(x) = 5^0 = 0

f(x) = 5^1 = 5

Use this formula to solve: f(b) - f(a)/b - a

5 - 0 / 1 - 0 = 5/1 = 5

--

f(x) = 5^2 = 25

f(x) = 5^3 = 125

125 - 25 / 3 - 2 = 100/1 = 100

Average rate of change: 

Section A: 5

Section B: 100

Part 2 (or b if you prefer): 

Section B is 20 times greater then A

This is because Section B is increasing in the equation. Thus, Section B is of greater value than Section A.

Hope this assists you and future students.

Answer:

288 hours.

Step-by-step explanation:

You exercise 24 hours in 1 month.

There are 12 months in one year.

24 * 12 = 288

Therefore, you exercised 288 hours in one year.

Answer:

Calculation with Distribution

24 (1+1+1+1+1+1+1+1+1+1+1+1)

24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1+24*1

24+24+24+24+24+24+24+24+24+24+24+24

288 hours

You did 288 hours of exercise.

Calculation without Distribution

24(1+1+1+1+1+1+1+1+1+1+1+1)

24(12)

288 hours

You did 288 hours of exercise.

Step-by-step explanation:

line AB the one marked c cm is:

8.2 * sin(49)=

6.1889

round answer to 6.2 cm

b. The slope is  - 400. This means that the helicopter descends 400 ft each minute. The slope is 400. This means that the helicopter ascends 400 ft each minute.

We know lines have various types of equations, the general type is

Ax + By + c = 0, and equation of a line in slope-intercept form is y = mx + b.

Where slope = m and b = y-intercept.

the slope is the rate of change of the y-axis with respect to the x-axis and the y-intercept is the (0,b) where the line intersects the y-axis at x = 0.

This situation is mathematically represented as a line and slope.

Points  (0, 10,000) and (15, 4000).

We know slope(m) = (y₂ - y₁)/(x₂ - x₁).

Slope(m) = (4000 - 10000)/(15 - 0).

Slope(m) = - 6000/15.

Slope(m) = - 400.

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An absolute value inequality is solved by re-writing it as compound inequality. For example.

|x+1| < 5

Since the value inside the absolute value brackets: (x+1) can be positive or negative, it is re-written as a compound inequality as the example below.

x+1 < 5

x+1 > - 5

solve for the range of values x can be

-6 < x < 4

Solving absolute value inequalities yields compound inequalities. Absolute value cases create intervals, linking the processes in algebraic solutions.

Solving absolute value inequalities and compound inequalities are related concepts in algebra, both involving multiple solutions and a range of possible values. Absolute value inequalities often result in compound inequalities due to the nature of the absolute value function.

When solving an absolute value inequality, such as [tex]\(|x - a| < b\), where \(a\)[/tex]and[tex]\(b\)[/tex] are constants, it generally leads to two separate cases: [tex]\(x - a < b\)[/tex] and [tex]\(x - a > -b\).[/tex] These cases represent the intervals where the expression inside the absolute value is less than [tex]\(b\)[/tex] and greater than its negative counterpart.

This process connects with solving compound inequalities, which involve combining two or more inequalities using logical connectors like "and" or "or." The solutions to the individual inequalities are then used to determine the overall solution to the compound inequality.

For instance, in the absolute value inequality[tex]\(|x - a| < b\)[/tex], the resulting compound inequality might be [tex]\(a - b < x < a + b\)[/tex]. This demonstrates that the solutions lie within the interval \((a - b, a + b)\), illustrating the linkage between solving absolute value inequalities and dealing with compound inequalities. Both concepts require careful consideration of multiple scenarios and the integration of solutions to find the overall solution set.

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