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ElenaW [278]
3 years ago
10

Translate the phrase '' Nine times the difference of a number and 3'' into an algebraic expression and then simplify ,Let x be r

epresent the real number....
Mathematics
1 answer:
Llana [10]3 years ago
8 0
Your question doesn't sound complete...the algebraic expression should equal something but this is what I ended up with 9(x+3)
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A chocolate chip cookie recipe calls for 2 cups of chocolate chips for every 3 cups of flour. How much flour would you use if yo
Blababa [14]

Answer:

you would use 1.5 cups of flour

4 0
3 years ago
How do I simplify −9(3−4w)−8w
laiz [17]
Multiply out the bracket so
-27 + 36w - 8w
Then collect the terms so
-27 + 28w

Answer = - 27 + 28w
3 0
2 years ago
Read 2 more answers
Solve the system of equations.<br><br><br><br> −2x+5y =−35<br> 7x+2y =25
Otrada [13]

Answer:

The equations have one solution at (5, -5).

Step-by-step explanation:

We are given a system of equations:

\displaystyle{\left \{ {{-2x+5y=-35} \atop {7x+2y=25}} \right.}

This system of equations can be solved in three different ways:

  1. Graphing the equations (method used)
  2. Substituting values into the equations
  3. Eliminating variables from the equations

<u>Graphing the Equations</u>

We need to solve each equation and place it in slope-intercept form first. Slope-intercept form is \text{y = mx + b}.

Equation 1 is -2x+5y = -35. We need to isolate y.

\displaystyle{-2x + 5y = -35}\\\\5y = 2x - 35\\\\\frac{5y}{5} = \frac{2x - 35}{5}\\\\y = \frac{2}{5}x - 7

Equation 1 is now y=\frac{2}{5}x-7.

Equation 2 also needs y to be isolated.

\displaystyle{7x+2y=25}\\\\2y=-7x+25\\\\\frac{2y}{2}=\frac{-7x+25}{2}\\\\y = -\frac{7}{2}x + \frac{25}{2}

Equation 2 is now y=-\frac{7}{2}x+\frac{25}{2}.

Now, we can graph both of these using a data table and plotting points on the graph. If the two lines intersect at a point, this is a solution for the system of equations.

The table below has unsolved y-values - we need to insert the value of x and solve for y and input these values in the table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & a \\ \cline{1-2} 1 & b \\ \cline{1-2} 2 & c \\ \cline{1-2} 3 & d \\ \cline{1-2} 4 & e \\ \cline{1-2} 5 & f \\ \cline{1-2} \end{array}

\bullet \ \text{For x = 0,}

\displaystyle{y = \frac{2}{5}(0) - 7}\\\\y = 0 - 7\\\\y = -7

\bullet \ \text{For x = 1,}

\displaystyle{y=\frac{2}{5}(1)-7}\\\\y=\frac{2}{5}-7\\\\y = -\frac{33}{5}

\bullet \ \text{For x = 2,}

\displaystyle{y=\frac{2}{5}(2)-7}\\\\y = \frac{4}{5}-7\\\\y = -\frac{31}{5}

\bullet \ \text{For x = 3,}

\displaystyle{y=\frac{2}{5}(3)-7}\\\\y= \frac{6}{5}-7\\\\y=-\frac{29}{5}

\bullet \ \text{For x = 4,}

\displaystyle{y=\frac{2}{5}(4)-7}\\\\y = \frac{8}{5}-7\\\\y=-\frac{27}{5}

\bullet \ \text{For x = 5,}

\displaystyle{y=\frac{2}{5}(5)-7}\\\\y=2-7\\\\y=-5

Now, we can place these values in our table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

As we can see in our table, the rate of decrease is -\frac{2}{5}. In case we need to determine more values, we can easily either replace x with a new value in the equation or just subtract -\frac{2}{5} from the previous value.

For Equation 2, we need to use the same process. Equation 2 has been resolved to be y=-\frac{7}{2}x+\frac{25}{2}. Therefore, we just use the same process as before to solve for the values.

\bullet \ \text{For x = 0,}

\displaystyle{y=-\frac{7}{2}(0)+\frac{25}{2}}\\\\y = 0 + \frac{25}{2}\\\\y = \frac{25}{2}

\bullet \ \text{For x = 1,}

\displaystyle{y=-\frac{7}{2}(1)+\frac{25}{2}}\\\\y = -\frac{7}{2} + \frac{25}{2}\\\\y = 9

\bullet \ \text{For x = 2,}

\displaystyle{y=-\frac{7}{2}(2)+\frac{25}{2}}\\\\y = -7+\frac{25}{2}\\\\y = \frac{11}{2}

\bullet \ \text{For x = 3,}

\displaystyle{y=-\frac{7}{2}(3)+\frac{25}{2}}\\\\y = -\frac{21}{2}+\frac{25}{2}\\\\y = 2

\bullet \ \text{For x = 4,}

\displaystyle{y=-\frac{7}{2}(4)+\frac{25}{2}}\\\\y=-14+\frac{25}{2}\\\\y = -\frac{3}{2}

\bullet \ \text{For x = 5,}

\displaystyle{y=-\frac{7}{2}(5)+\frac{25}{2}}\\\\y = -\frac{35}{2}+\frac{25}{2}\\\\y = -5

And now, we place these values into the table.

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

When we compare our two tables, we can see that we have one similarity - the points are the same at x = 5.

Equation 1                  Equation 2

\begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & -7 \\ \cline{1-2} 1 & -33/5 \\ \cline{1-2} 2 & -31/5 \\ \cline{1-2} 3 & -29/5 \\ \cline{1-2} 4 & -27/5 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}                 \begin{array}{|c|c|} \cline{1-2} \textbf{x} & \textbf{y} \\ \cline{1-2} 0 & 25/2 \\ \cline{1-2} 1 & 9 \\ \cline{1-2} 2 & 11/2 \\ \cline{1-2} 3 & 2 \\ \cline{1-2} 4 & -3/2 \\ \cline{1-2} 5 & -5 \\ \cline{1-2} \end{array}

Therefore, using this data, we have one solution at (5, -5).

4 0
3 years ago
Factorise each of the following algebraic expressions completely,
LekaFEV [45]

Answer:

see explanation

Step-by-step explanation:

(a)

Given

2k - 6k² + 4k³ ← factor out 2k from each term

= 2k(1 - 3k + 2k²)

To factor the quadratic

Consider the factors of the product of the constant term ( 1) and the coefficient of the k² term (+ 2) which sum to give the coefficient of the k- term (- 3)

The factors are - 1 and - 2

Use these factors to split the k- term

1 - k - 2k + 2k² ( factor the first/second and third/fourth terms )

1(1 - k) - 2k(1 - k) ← factor out (1 - k) from each term

= (1 - k)(1 - 2k)

1 - 3k + 2k² = (1 - k)(1 - 2k) and

2k - 6k² + 4k³ = 2k(1 - k)(1 - 2k)

(b)

Given

2ax - 4ay + 3bx - 6by ( factor the first/second and third/fourth terms )

= 2a(x - 2y) + 3b(x - 2y) ← factor out (x - 2y) from each term

= (x - 2y)(2a + 3b)

8 0
3 years ago
20. Based on student records, 25% of the students at a large high school have a
Elenna [48]

Answer:

(E) 0.71

Step-by-step explanation:

Let's call A the event that a student has GPA of 3.5 or better, A' the event that a student has GPA lower than 3.5, B the event that a student is enrolled in at least one AP class and B' the event that a student is not taking any AP class.

So, the  probability that the student has a GPA lower than 3.5 and is not taking any AP  classes is calculated as:

P(A'∩B') = 1 - P(A∪B)

it means that the students that have a GPA lower than 3.5 and are not taking any AP  classes are the complement of the students that have a GPA of 3.5 of better or are enrolled in at least one AP class.

Therefore, P(A∪B) is equal to:

P(A∪B) = P(A) + P(B) - P(A∩B)

Where the probability P(A) that a student has GPA of 3.5 or better is 0.25, the probability P(B) that a student is enrolled in at least one AP class is 0.16 and the probability P(A∩B) that a student has a GPA of 3.5 or better and is enrolled  in at least one AP class is 0.12

So, P(A∪B) is equal to:

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∪B) = 0.25 + 0.16 - 0.12

P(A∪B) = 0.29

Finally, P(A'∩B') is equal to:

P(A'∩B') = 1 - P(A∪B)

P(A'∩B') = 1 - 0.29

P(A'∩B') = 0.71

5 0
3 years ago
Read 2 more answers
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