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scoray [572]
3 years ago
7

Question 13) Olivia has read 143 pages of a 220 page novel. What

Mathematics
2 answers:
igomit [66]3 years ago
7 0

Answer:

77%

Step-by-step explanation:

Reil [10]3 years ago
4 0

Answer:

The answer to Question 13 is 35%.

Step-by-step explanation:

Subtract 143 from 220, which will give you 77. Then, you take that 77 and search, "77 out of 220" in Google. You should get 35 and just put the percentage symbol. Olivia still needs to read 35% of the 220 page book:).

You might be interested in
PROMPT:
pentagon [3]
For this answer let's pick the Fish Pond.

QUESTION: <span>What do you know? 
We know that the total area of the rectangular patio is 24 square feet. The rectangular fish pond that we want to build should have the length that is twice the width. We also need a 1-foot border around the pond.

The total area of the patio is represented by </span>2 x^{2} +6x+4 where x is the width of the fish pond.

QUESTION: What do you want to find out? 
We can answer this question by examining what we have been given (we have answered this in the previous question). Since we only have one variable in the equation and this is the width of the fish pond, then this is the only thing that we need to find out. The length will just follow after the width is solved. (Basically, we are concerned with the dimensions of our project.)

QUESTION: What kind of answer do you expect? 
Since we are only expecting to solve for the dimensions, then we should expect two numbers which would represent the width and the length. Through the polynomial, we can get the width of our project, then the length would just be twice our answer for the width.

1. To set our polynomial equation equal to the total area of our patio, we just write the equation on the left side and the total area of the patio on the right. We recall the equation that we gave in the first question above as well as the patio's area.

2x^{2} +6x+4=24

2. For this item, we just simply negate the constant term on the right side by subtracting the same number to both sides. Since our constant term is 24, we subtract 24 from the polynomial as well as the term on the right side.

2x^{2} +6x+4-24=24-24
2x^{2} +6x-20=0

3. In this item we are only tasked to find the GCF of the trinomial and factor it out of the left side of the equation. The GCF just means the common number or variable in all terms. Upon examining, we can see that the only common factor of the terms is 2 (since all terms are divisible by 2), thus this will be the GCF and we will factor this one out.

2(x^{2} +3x-10)=0

4. To factor the polynomial completely, we just figure out the possible factors of the quadratic equation. You can do this by trying out all factors of -10 that will have a sum of 3 (since -10 is the constant and 3 is the coefficient of x). Upon examining you should end up with:

2(x+5)(x-2)=0 (Notice that 5 and -2 are factors of -10 and their sum is 3).

5. The rule stated on this item just means that we can equate each individual factor to zero (except for the constant term) to find out the possible values of x. That means that we can solve for x by using the equations x+5=0 and x-2=0.

x+5=0
x=-5

x-2=0
x=2

6. For this item, we just do simple substitutions to the equation 2x^{2} +6x-20=0 to verify whether the values we got in question 5 is really a solution to the equation. The values of x that we got are -5 and 2 so we substitute these one at a time.

2(-5)^{2} +6(-5)-20=0
50-30-20=0
0=0

2(2)^{2} +6(2)-20=0
8+12-20=0
0=0

Both values of x make the left side equal to zero therefore both are solutions to the equation.

7. We can tell the dimensions of the project by looking at the values that we got for x, since we assumed x to be the width of our project. The fact that we got two values for x won't be a problem since one of these values is negative. There is no negative measurement/width so we can just ignore this negative value (-5). Thus, the width for our project would be 2 and the length would be twice this value which is 4.

Width = 2 feet
Length = 4 feet

8. For this item, we just illustrate the dimensions of our project and add a 1-foot border (as we are told initially). You can see this illustration in the picture I attached below. 

Since we have an extra 1 foot in every side, that means we need to add 2 feet to both the width and the length. Therefore, our width now is 4 while our length is 6.

Multiplying these numbers to verify the area, we get 24 square feet which is exactly the area of the patio.

6 0
3 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
Describe the solutions of 5 &lt;n​
Ainat [17]

Answer:

D

Step-by-step explanation:

All real number less than 5

4 0
4 years ago
May someone please help me with this :)
Studentka2010 [4]

Answer:

i was taught this but i dont remember so I dont know if I can help sorry bud

8 0
3 years ago
M(4,2) is the midpoint of RS. The coordinates of S are (6, 1). what are the coordinates of R?
nirvana33 [79]
Remark
You are using the midpoint formula. Instead of finding the midpoint, you are looking for one of the points, so you have to rearrange the formula a little bit.

Givens
Midpoint (4,2)
One endpoint (6,1)

Object
Find the other endpoint.

Formula
m(x,y) = (x1 + x2)/2, (y1 + y2)/2)

Solution
Find the x value
4 = (6 + x2)/2    Multiply both sides by 2
4*2 = 6 + x2      Subtract 6 from both sides.
8 - 6 = x2
x2 = 2

Find the y value
2 = (1 + y2)/2    Multiply by 2
4 = 1 + y2          Subtract 1 from both sides.
4 - 1 = y2  
y2 = 3

Conclusion
R(x,y) = (2,3)

8 0
3 years ago
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