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kakasveta [241]
3 years ago
6

Carlos pays for dinner with a friend. He uses a coupon for $5 off the price of dinner, before taxes are applied. The tax rate is

8%. The total cost for dinner after the coupon and taxes is $30.78. What is the original price of dinner before the coupon and taxes.
Mathematics
1 answer:
Lostsunrise [7]3 years ago
5 0

Now after the coupon of $5.00 and after paying the taxes at the rate of 8%. He paid $30.78 for the dinner.

So, lets say the actual price of the dinner was 'x' dollars.

Now, 8% of x dollars can be written as,

\frac{8}{100} \timesx=0.08x, this is the amount he needs to pay in addition to the original price (since it is taxation on the actual price).

And he got off of $5.00 from the coupon.

So after all the deductions and payment of tax, he paid $30.78, which can also be written as:

x+0.08x-5=30.78

Solving for 'x' we get:

1.08x-5=30.78

1.08x=30.78+5

1.08x=35.78

x=\frac{35.78}{1.08} =33.129

Therefore, the actual cost of the dinner is $33.13.

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Solve for n
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What is the area of this polygon in square units ?
Maru [420]

Answer:

9 units²

Step-by-step explanation:

This is a trapezium.

The length of the two parallel lines (on top and bottom) are 5 units and 1 units.

The height of the trapezium is 3 units. Using these information, we can apply the formula for area of trapezium to find the answer.

Area = 1/2 (5 + 1) x 3 = 9 units²

7 0
3 years ago
Without plotting points, let M=(-2,-1), N=(3,1), M'= (0,2), and N'=(5, 4). Without using the distanceformula, show that segments
kramer

Given:

M=(x1, y1)=(-2,-1),

N=(x2, y2)=(3,1),

M'=(x3, y3)= (0,2),

N'=(x4, y4)=(5, 4).

We can prove MN and M'N' have the same length by proving that the points form the vertices of a parallelogram.

For a parallelogram, opposite sides are equal

If we prove that the quadrilateral MNN'M' forms a parallellogram, then MN and M'N' will be the oppposite sides. So, we can prove that MN=M'N'.

To prove MNN'M' is a parallelogram, we have to first prove that two pairs of opposite sides are parallel,

Slope of MN= Slope of M'N'.

Slope of MM'=NN'.

\begin{gathered} \text{Slope of MN=}\frac{y2-y1}{x2-x1} \\ =\frac{1-(-1)}{3-(-2)} \\ =\frac{2}{5} \\ \text{Slope of M'N'=}\frac{y4-y3}{x4-x3} \\ =\frac{4-2}{5-0} \\ =\frac{2}{5} \end{gathered}

Hence, slope of MN=Slope of M'N' and therefore, MN parallel to M'N'

\begin{gathered} \text{Slope of MM'=}\frac{y3-y1}{x3-x1} \\ =\frac{4-(-1)}{5-(-2)} \\ =\frac{3}{2} \\ \text{Slope of NN'=}\frac{y4-y2}{x4-x2} \\ =\frac{4-1}{5-3} \\ =\frac{3}{2} \end{gathered}

Hence, slope of MM'=Slope of NN' nd therefore, MM' parallel to NN'.

Since both pairs of opposite sides of MNN'M' are parallel, MM'N'N is a parallelogram.

Since the opposite sides are of equal length in a parallelogram, it is proved that segments MN and M'N' have the same length.

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1 year ago
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