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3 years ago

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Leyan bought 234 pounds of apples for $1.32 per pound, 112 pounds of peaches for $1.20 per pound, and some tomatoes, all from a

grocery store. The total cost of his purchase was $9.63. How much money did Leyan spend on tomatoes?

2 answers:

velikii [3]3 years ago

8 0

Answer: See explanation

Step-by-step explanation:

I think your question isn't well written and that 234 was meant to be 2.34 and 112 was meant to be 1.12

Let the amount spent on tomatoes be x.

The information given in the question can then be used to form an equation as thus:

Apples + Peaches + Tomatoes = $9.63

(2.34 × $1.32) + (1.12 × $1.20) + x = $9.63

$3.0888 + $1.344 + x = $9.63

x = $9.63 - ($3.0888 + $1.344)

x = $9.63 - $4.4328

x = $5.1972

The amount spent on tomatoes will be $5.1972

telo118 [61]3 years ago

5 0

Answer:

The answer is 4.2 I did this question already

Step-by-step explanation: You multiply 2 3/4 and 1.32 and then add that to the product of the 1 1/2 peaches for 1.20 per pound, then subtract the answer from 9.63 to get the solution for x.

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(03.05 MC) Solve the rational equation x divided by 2 equals x squared divided by quantity x minus 2 end quantity, and check for

ycow [4]

Answer:

x = 0 and x = -2 are solutions of the given rational equation.

Step-by-step explanation:

We must solve the following rational equation:

$\frac{x}{2} = \frac{x^{2}}{x-2}$

Now we present the procedure:

1) $\frac{x}{2} = \frac{x^{2}}{x-2}$ Given

2) $x\cdot (x-2) = 2\cdot x^{2}$ Compatibility with multiplication/Existence of the multiplicative inverse/Definition of division/Modulative property.

3) $x^{2}-2\cdot x = 2\cdot x^{2}$ Distributive property/ $a^{b}\cdot a^{c} = a^{b+c}$

4) $x^{2} + 2\cdot x = 0$ Compatibility with addition/Existence of the additive inverse/Modulative property/Reflexive property

5) $x \cdot (x+2) = 0$ Distributive property/ $a^{b}\cdot a^{c} = a^{b+c}$

6) $x = 0\, \lor\, x = -2$ Result

Now we check the rational equation with each root:

x = 0

$\frac{x}{2} = \frac{x^{2}}{x-2}$

$\frac{0}{2} = \frac{0^{2}}{0-2}$

$0 = \frac{0}{-2}$

$0 = 0$

x = 0 is a solution of the rational equation.

x = -2

$\frac{x}{2} = \frac{x^{2}}{x-2}$

$\frac{-2}{2} = \frac{(-2)^{2}}{-2-2}$

$-1 = -1$

x = -2 is a solution of the rational equation.

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3 years ago

A line passes through the points (-7, 2) and (1, 6).A second line passes through the points (-3, -5) and (2, 5).Will these two l

BlackZzzverrR [31]

Answer:

Yes, the lines intersect at (3,7). The solution is (3,7).

Explanation:

Step 1. The first line passes through the points:

(-7,2) and (1,6)

and the second line passes through the points:

(-3,-5) and (2,5)

Required: State if the lines intersect, and if so, find the solution.

Step 2. We need to find the slope of the lines.

Let m1 be the slope of the first line and m2 be the slope of the second line.

The formula to find a slope when given two points (x1,y1) and (x2,y2) is:

$m=\frac{y_2-y_1}{x_2-x_1}$

Using our two points for each line, their slopes are:

$\begin{gathered} m_1=\frac{6-2}{1-(-7)} \\ \\ m_2=\frac{5-(-5)}{2-(-3)} \end{gathered}$

The results are:

$\begin{gathered} m_1=\frac{6-2}{1-(-7)}=\frac{4}{1+7}=\frac{4}{8}=\frac{1}{2} \\ \\ \end{gathered}$ $m_2=\frac{5+5}{2+3}=\frac{10}{5}=2$

The slopes are not equal, this means that the lines are NOT parallel, and they will intersect at some point.

Step 3. To find the intersection point (the solution), we need to find the equation for the two lines.

Using the slope-point equation:

$y=m(x-x_1)+y_1$

Where m is the slope, and (x1,y1) is a point on the line.

For the first line m=1/2, and (x1,y1) is (-7,2). The equation is:

$y=\frac{1}{2}(x-(-7))+2$

Solving the operations:

$\begin{gathered} y=\frac{1}{2}(x+7)+2 \\ \downarrow\downarrow \\ y=\frac{1}{2}x+7/2+2 \\ \downarrow\downarrow \\ y=\frac{1}{2}x+5.5 \end{gathered}$

Step 4. We do the same for the second line. The slope is 2. and the point (x1,y1) is (-3, -5). The equation is:

$\begin{gathered} y=2(x-(-3))-5 \\ \downarrow\downarrow \\ y=2x+6-5 \\ \downarrow\downarrow \\ y=2x+1 \end{gathered}$

Step 5. The two equations are:

$\begin{gathered} y=\frac{1}{2}x+5.5 \\ y=2x+1 \end{gathered}$

Now we need to solve for x and y.

Step 6. Equal the two equations to each other:

$\frac{1}{2}x+5.5=2x+1$

And solve for x:

$\begin{gathered} \frac{1}{2}x+5.5=2x+1 \\ \downarrow\downarrow \\ 5.5-1=2x-\frac{1}{2}x \\ \downarrow\downarrow \\ 4.5=1.5x \\ \downarrow\downarrow \\ \frac{4.5}{1.5}=x \\ \downarrow\downarrow \\ \boxed{3=x} \end{gathered}$

Step 7. Use the second equation:

$y=2x+1$

and substitute the value of x to find the value of y:

$\begin{gathered} y=2(3)+1 \\ \downarrow\downarrow \\ y=6+1 \\ \downarrow\downarrow \\ \boxed{y=7} \end{gathered}$

The solution is x=3 and y=7, in the form (x,y) the solution is (3,7).

Answer:

Yes, the lines intersect at (3,7). The solution is (3,7).

6 0

1 year ago