The fourth grade students at Riverside school are going on a field trip. There are 68 students on each of the 4 buses. How many students are going on the field trip?

Let

x------> the number of men’s shirts

y----->  the number of women’s shirts

we know that

[tex]8x+12y \leq 240[/tex] ----> constraint A    

[tex]x\geq 3[/tex] ----> constraint B

[tex]y\geq 8[/tex] ----> constraint C  

[tex]y\geq 2x[/tex] ----> constraint D  

using a graphing tool

see the attached figure N[tex]1[/tex]

The solution is the shaded area

Remember that

if an ordered pair meets all restrictions and makes sense in the context of the situation, then the ordered pair is a system solution

we're going to proceed to verify each pair ordered

case A) [tex](4,9)[/tex]

Substitute the value of x and y in each  constraint

[tex]x=4\\y=9[/tex]

Verify constraint A

[tex]8*4+12*9 \leq 240[/tex]

[tex]140 \leq 240[/tex] -------> is true

Verify constraint B

[tex]4\geq 3[/tex] -------> is true

Verify constraint C

[tex]9\geq 8[/tex] ------> is true

Verify constraint D

[tex]9\geq 2*4[/tex]

[tex]9\geq 8[/tex] -----> is true

therefore

the ordered pair [tex](4,9)[/tex]  is a system solution

case B) [tex](7,13)[/tex]

Substitute the value of x and y in each  constraint

[tex]x=7\\y=13[/tex]

Verify constraint A      

[tex]8*7+12*13 \leq 240[/tex]

[tex]212 \leq 240[/tex] -------> is true

Verify constraint B

[tex]7\geq 3[/tex] -------> is true

Verify constraint C

[tex]13\geq 8[/tex] ------> is true

Verify constraint D

[tex]13\geq 2*7[/tex]

[tex]13\geq 14[/tex] -----> is not true

therefore

the ordered pair [tex](7,13)[/tex] is not a system solution

case C) [tex](2,15)[/tex]

Substitute the value of x and y in each  constraint

[tex]x=2\\y=15[/tex]

Verify constraint A

[tex]8*2+12*15 \leq 240[/tex]

[tex]196 \leq 240[/tex] -------> is true

Verify constraint B

[tex]2\geq 3[/tex] -------> is not true

Verify constraint C

[tex]15\geq 8[/tex] ------> is true

Verify constraint D

[tex]15\geq 2*2[/tex]

[tex]15\geq 4[/tex] -----> is true

therefore

the ordered pair [tex](2,15)[/tex] is not a system solution  

case D) [tex](5,12)[/tex]

Substitute the value of x and y in each  constraint

[tex]x=5\\y=12[/tex]

Verify constraint A

[tex]8*5+12*12 \leq 240[/tex]

[tex]184 \leq 240[/tex] -------> is true

Verify constraint B

[tex]5\geq 3[/tex] -------> is true

Verify constraint C

[tex]12\geq 8[/tex] ------> is true

Verify constraint D

[tex]12\geq 2*5[/tex]

[tex]12\geq 10[/tex] -----> is true

therefore

the ordered pair [tex](5,12)[/tex] is  a system solution  

case E) [tex](6,15)[/tex]

Substitute the value of x and y in each  constraint

[tex]x=6\\y=15[/tex]

Verify constraint A

[tex]8*6+12*15 \leq 240[/tex]

[tex]228 \leq 240[/tex] -------> is true

Verify constraint B

[tex]6\geq 3[/tex] -------> is true

Verify constraint C

[tex]15\geq 8[/tex] ------> is true

Verify constraint D

[tex]15\geq 2*6[/tex]

[tex]15\geq 12[/tex] -----> is true

therefore

the ordered pair [tex](6,15)[/tex] is  a system solution  

therefore

the answer is

[tex](4,9)[/tex]

[tex](5,12)[/tex]

[tex](6,15)[/tex]

see the attached figure N[tex]2[/tex] to better understand the solution

Answer:

(4,9) (5,12) (6,15)

Step-by-step explanation:

Those were the correct options.

Answer:

The graph in the attached figure

Step-by-step explanation:

Let

x------> the number of hammers

y------> the number of of wrenches

we know that

[tex]10x+6y\leq 120[/tex] -----> inequality A

[tex]x+y\geq 14[/tex] -----> inequality B    

using a graphing tool

The solution of the system of inequalities is the shaded area  

see the attached figure

The number is 30+ 3x².

When an integer is multiplied by itself, the resultant number is known as its square number. Basically, a square number is a number that is obtained by the product of two same numbers. In geometry, the area of a square with 'n' as the side length (where n is an integer) is the finest example of a square number.

When an integer is multiplied by the same integer, the resultant number is known as a square number. For example, when we multiply 5 × 5 = 52, we get 25. Here, 25 is a square number. Square numbers are always positive, they cannot be negative because when a negative number is multiplied by the same negative number, it results in a positive number.

let the number be x

Now, 3 times square of a number

3x²

and, 30 increased

=30+ 3x²

Hence, the number is 30+ 3x².

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The equation 2m - nx = x + 4 is a linear equation solvable for x using algebraic manipulation. The solution is x = (2m - 4)/(nx + 1), requiring values for m and n to find x's specific value.

The given equation 2m - nx = x + 4 is a linear equation and can be solved for x by applying algebraic techniques. The question indicates that there is an element of differential equations involved, suggesting that it might be related to more advanced topics such as linear differential equations or characteristics equations. However, the provided context also mentions concepts like probability with positive and negative jumps, quadratic equations, and principles from physics, but these do not directly apply to the initial linear equation provided.

To solve for x in the given equation, we would rearrange the terms to isolate x:

2m - nx = x + 4

2m - 4 = x(nx + 1)

x = (2m - 4)/(nx + 1)

The solution for x is now expressed in terms of m and n. To find the specific value of x, values for m and n must be known.

False, because the difference between two negative numbers is not always negative.

Here,

Given that, The difference between two negative numbers is always negative.

We have to prove this statement is true or false.

What is Negative number?

In the real number system, a negative number is a number that is less than zero.

Now,

We can prove it false with a counter example.

Let two negative number -6 and -8.

Hence, difference between -6 and -8 is,

⇒-6 - ( -8 )

⇒-6 + 8

⇒2

it is not negative number.

Hence, the difference between two negative numbers is not always negative.

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Final answer:

To round 5,503,569 to the nearest hundred, observe the hundreds and the following digit. Since the following digit is a 9, round the hundreds digit up to 7 and zeros follow. The rounded number is 5,503,600.

Explanation:

To round the number 5,503,569 to the nearest hundred, we need to look at the digit in the hundreds place and the digit following it. In 5,503,569, the hundreds digit is 6, and the digit to its right is 9. According to rounding rules, if the digit right after the one we are rounding is 5 or greater, we round up. Since 9 is greater than 5, we round the hundreds place up from 6 to 7 and change all the digits to the right of the hundreds place to zero.

The final answer, therefore, is 5,503,600 when 5,503,569 is rounded to the nearest hundred.

Answer:

[tex]y=m^2-4m-45[/tex]

Step-by-step explanation:

An equation has solutions of m = –5 and m = 9

WE are given with the solution. Lets write the solution as factors

When x=a is a solution then factor is (x-a)

[tex]m=-5[/tex] is a solution. change the sign of the solution while writing factor. factor is (m+5)

[tex]m=9[/tex] is a solution, factor is (m-9)

we use the factors to find the equation

[tex]y=(m+5)(m-9)[/tex]

Multiply the factors using FOIL method

[tex]y=m^2-9m+5m-45[/tex]

[tex]y=m^2-4m-45[/tex]

Answer: 0.2

Step-by-step explanation:

To calculate the distance that Joanie run during her lunch break is necessary to multiply the total distance around the outside of Cedar Park (0,8) by the 1/4 of the total distance that she ran (0,25).

So, the expression is:

0,25 . 0.8 = 0,2

Joanie run 0,2 miles during  her  lunch break.

Answer:

option B

Step-by-step explanation:

We have three different function for h(x)

[tex]h(x)= \frac{1}{4}x-4, x<=0[/tex]

WE have x<=0, so the graph starts at x=0 and goes down

Plug in 0 for x

[tex]h(0)= \frac{1}{4}(0)-4[/tex]

So h(0)= -4

When x=0, y= -4

The graph starts at (0,-4) and goes down

[tex]h(x)= \frac{1}{3}x-3, 0<x<=3[/tex]

x lies between 0  and 3, so the graph starts at x=0 and ends at x=3

Plug in 0 for x

[tex]h(0)= \frac{1}{3}(0)-3[/tex]

So h(0)= -3

When x=0, y= -4

Plug in 3 for x

[tex]h(3)= \frac{1}{3}(3)-3=-2[/tex]

When x=3, y= -2

The graph starts at (0,-4) and ends at (3, -2)

[tex]h(x)= \frac{1}{2}x-2, x>=4[/tex]

WE have x>=4, so the graph starts at x=4 and goes to the right

Plug in 4 for x

[tex]h(4)= \frac{1}{2}(4)-2[/tex]

So h(4)= 0

When x=4, y= 0

The graph starts at (4,0) and goes to the right.

SEcond graph is correct

Answer:

B

Step-by-step explanation:

Answer:

[tex]\dfrac{2x^2}{x} + x(100-15x)[/tex] at x=5

135

Step-by-step explanation:

Given: The expression [tex]\dfrac{2x^2}{x} + x(100-15x)[/tex]

This is rational expression of variable x.

Put x=5 into the expression and then simplify

[tex]\Rightarrow \dfrac{2\cdot 5^2}{5} + 5(100-15\cdot 5)[/tex]

[tex]\Rightarrow 2\cdot 5 + 5(100-75)[/tex]

[tex]\Rightarrow 2\cdot 5 + 5\cdot 25)[/tex]

[tex]\Rightarrow 10+125[/tex]         (Addition of two integer)

[tex]\Rightarrow 135[/tex]

Hence, The value of given expression is 135

Answer:

Step-by-step explanation:

You want the reasons for each of the final three steps in the solution of the inequality 7x +16 < -5(5 -5x).

2. 7x +16 < -20 +25x . . . . . . . given step 2

3.  16 < -20 +18x . . . . . Subtraction property of inequality

4.  36 < 18x . . . . . . . . . Addition property of inequality

5.  2 < x . . . . . . . . . . . . Division property of inequality

__

Additional comment

Subtraction of 7x is the same as adding -7x, so you can say the reason for step 3 is either of "subtraction property" or "addition property" The same would be true of step 4.

To figure this out, you compare the inequality on each line with the one before, and determine what was changed. If the same thing is done to both sides of the inequality (addition, subtraction, multiplication, division), the reason is the corresponding property of inequality.

The multiplication and division properties of inequality tell you to reverse the comparison symbol when the factor used is negative. Here, the factor is 18, a positive number, so the inequality symbol remains untouched.

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The rate of change of the height of the water in the pool when the depth is 5 feet is approximately 0.089 feet per minute.

To find the rate of change of the height of the water in the pool, we can use the formula for the volume of a cylinder.

The volume of a cylinder is given by the formula V = πr^2h, where r is the radius and h is the height.

So, when the depth of the water is 5 feet, we can find the rate of change of the height by differentiating the volume equation with respect to time.

Let's calculate it:

V = π(5^2)(h)

dV/dt = π(25)(dh/dt)

Since the volume is increasing at a constant rate of 7 cubic feet per minute, we have dV/dt = 7.

Substituting the given values, we have:

7 = π(25)(dh/dt)

dh/dt = 7/π(25)

So, the rate of change of the height of the water in the pool when the depth is 5 feet is approximately 0.089 feet per minute.

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To find the rate of change of the height of the water in the pool, we can use the concept of related rates. The rate of change is approximately 0.089 cubic feet per minute.

To find the rate of change of the height of the water in the pool, we can use the concept of related rates. Let's call the height of the water in the pool 'h' and the rate of change of the height 'dh/dt'. We know that the volume of water in the pool is flowing at a constant rate of 7 cubic feet per minute, therefore the rate of change of the volume of water in the pool is also constant at 7 cubic feet per minute. The volume of a cylinder is given by V = πr^2h, where r is the radius of the pool and h is the height of the water. We can differentiate this equation with respect to time to find the rate of change of the volume.

dV/dt = πr^2(dh/dt)

Since the radius of the pool is constant at 5 feet and the rate of change of the volume is 7 cubic feet per minute, we can substitute these values into the equation and solve for dh/dt.

7 = π(5^2)(dh/dt)

dh/dt = 7/(π(5^2))

Therefore, the rate of change of the height of the water in the pool when the depth of the water is 5 feet is approximately 0.089 cubic feet per minute.

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