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2 years ago

Martha needs 7 strawberries for every 1 smoothie. What are the missing values in the table

Mathematics

2 answers:

Ede4ka [16]2 years ago

7 0

Answer:

Strawberries Smoothies

28 4

21 3

70 10

Step-by-step explaination

If Maria uses 28 strawberries you would divide that by 7 (since that's how many she uses for one smoothie) after you divide 28 by 7 you get 4, 4 is the number of smoothies 28 strawberries makes.

If Maria makes 3 smoothies you multiple 3 by 7 (since that's how many strawberries she uses to make one) 7 times 3 is 21. If you want to check your work you would divide 21 by 7 (the number of strawberries used for 1 smoothie) and you would get 3.

If Maria uses 70 strawberries you would divide 70 by 7 (the number of strawberries used for 1 smoothie) 70 divided by 7 is 10, 10 is the number of smoothies 70 strawberries make.

kodGreya [7K]2 years ago

4 0

4, 21, 10
————
28-4
21-3
70-10

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================================================

Work Shown:

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3 years ago

Read 2 more answers

X/5-2=x/2+3 solve for x

Makovka662 [10]

Solve for x:x/5 - 2 = x/2 + 3
Put each term in x/5 - 2 over the common denominator 5: x/5 - 2 = x/5 - (10)/5:x/5 - (10)/5 = x/2 + 3
x/5 - (10)/5 = (x - 10)/5:(x - 10)/5 = x/2 + 3
Put each term in x/2 + 3 over the common denominator 2: x/2 + 3 = x/2 + 6/2:(x - 10)/5 = x/2 + 6/2
x/2 + 6/2 = (x + 6)/2:(x - 10)/5 = (x + 6)/2
Multiply both sides by 10:(10 (x - 10))/5 = (10 (x + 6))/2
10/5 = (5×2)/5 = 2:2 (x - 10) = (10 (x + 6))/2
10/2 = (2×5)/2 = 5:2 (x - 10) = 5 (x + 6)
Expand out terms of the left hand side:2 x - 20 = 5 (x + 6)
Expand out terms of the right hand side:2 x - 20 = 5 x + 30
Subtract 5 x from both sides:(2 x - 5 x) - 20 = (5 x - 5 x) + 30
2 x - 5 x = -3 x:-3 x - 20 = (5 x - 5 x) + 30
5 x - 5 x = 0:-3 x - 20 = 30
Add 20 to both sides:(20 - 20) - 3 x = 20 + 30
20 - 20 = 0:-3 x = 30 + 20
30 + 20 = 50:-3 x = 50
Divide both sides of -3 x = 50 by -3:(-3 x)/(-3) = 50/(-3)
(-3)/(-3) = 1:x = 50/(-3)
Multiply numerator and denominator of 50/(-3) by -1:Answer: x = (-50)/3

7 0

2 years ago

Solve the system of equations. −2x+5y =−35 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:

Graphing the equations (method used)
Substituting values into the equations
Eliminating variables from the equations

Graphing the Equations

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.

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

2 years ago