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Mnenie [13.5K]
4 years ago
14

Hiroto paid $4.28 in sales tax for the items that he purchased. If the sales tax rate was 6.25 percent, what was the total cost

of the items before tax was added?
Mathematics
1 answer:
ELEN [110]4 years ago
5 0

Answer:

$68.48

Step-by-step explanation:

Using a chart....

100% divided by 6.25 percent gives us 16 columns

so 16 columns of 6.25 % to equal 100%

second row is 4.28 each

4.28 times 16 = $68.48

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Im stuck on this question, plz help
denis-greek [22]

Answer:

Kindly check explanation

Step-by-step explanation:

Given the following :

40 metres covered in 6 seconds

Max speed attained after 6seconds

75 meters covered after 11 seconds

Average Speed for first 40 meters :

Speed = distance / time

Speed = 40m / 6s

Speed = 6.67m/s

To obtain the maximum speed :

Next (75 - 40) meters = 35 meters was covered in (11 - 6)seconds = 5 seconds

Speed at this point is maximum :

Hence, maximum speed = (35m / 5s) = 7m/s

Suppose, Manuel runs for an additional z seconds after reaching max speed :

Distance from starting line 6+z seconds after race started?

Distance after 6 seconds = 40 metres

Distance after z seconds = 7 * z

Total distance = (40 + 7z)

What is Manuel's distance from the starting line x seconds after the race started (provided x≥6x)?

Distance for first 6 seconds = 40 meters + distance covered after 6 seconds = (7 * (x-6))

40 + 7(x - 6)

3 0
3 years ago
Write each rate as a unit rate. $2 for 5 cans of soup.
Len [333]
If you would like to write each rate as a unit rate, you can do this using the following steps:

$2 ... 5 cans of a soup
$1 ... x cans of a soup = ? 

2 * x = 5 * 1
2 * x = 5
x = 5 / 2
x = 2.5 cans of a soup per $1

$2 ... 5 cans of a soup
$x = ? ... 1 can of a soup

2 * 1 = 5 * x
2 = 5 * x
x = 2 / 5 
x = $0.4 per 1 can of a soup
6 0
3 years ago
Please help me!!!!!!
Genrish500 [490]
Aye do 9*9*6 then it gives u the answer

8 0
3 years ago
Evaluate the line integral by the two following methods. xy dx + x2 dy C is counterclockwise around the rectangle with vertices
Airida [17]

Answer:

25/2

Step-by-step explanation:

Recall that for a parametrized differentiable curve C = (x(t), y(t)) with the parameter t varying on some interval [a, b]

\large \displaystyle\int_{C}[P(x,y)dx+Q(x,y)dy]=\displaystyle\int_{a}^{b}[P(x(t),y(t))x'(t)+Q(x(t),y(t))y'(t)]dt

Where P, Q are scalar functions

We want to compute

\large \displaystyle\int_{C}P(x,y)dx+Q(x,y)dy=\displaystyle\int_{C}xydx+x^2dy

Where C is the rectangle with vertices (0, 0), (5, 0), (5, 1), (0, 1) going counterclockwise.

a) Directly

Let us break down C into 4 paths \large C_1,C_2,C_3,C_4 which represents the sides of the rectangle.

\large C_1 is the line segment from (0,0) to (5,0)

\large C_2 is the line segment from (5,0) to (5,1)

\large C_3 is the line segment from (5,1) to (0,1)

\large C_4 is the line segment from (0,1) to (0,0)

Then

\large \displaystyle\int_{C}=\displaystyle\int_{C_1}+\displaystyle\int_{C_2}+\displaystyle\int_{C_3}+\displaystyle\int_{C_4}

Given 2 points P, Q we can always parametrize the line segment from P to Q with

r(t) = tQ + (1-t)P for 0≤ t≤ 1

Let us compute the first integral. We parametrize \large C_1 as

r(t) = t(5,0)+(1-t)(0,0) = (5t, 0) for 0≤ t≤ 1 and

r'(t) = (5,0) so

\large \displaystyle\int_{C_1}xydx+x^2dy=0

 Now the second integral. We parametrize \large C_2 as

r(t) = t(5,1)+(1-t)(5,0) = (5 , t) for 0≤ t≤ 1 and

r'(t) = (0,1) so

\large \displaystyle\int_{C_2}xydx+x^2dy=\displaystyle\int_{0}^{1}25dt=25

The third integral. We parametrize \large C_3 as

r(t) = t(0,1)+(1-t)(5,1) = (5-5t, 1) for 0≤ t≤ 1 and

r'(t) = (-5,0) so

\large \displaystyle\int_{C_3}xydx+x^2dy=\displaystyle\int_{0}^{1}(5-5t)(-5)dt=-25\displaystyle\int_{0}^{1}dt+25\displaystyle\int_{0}^{1}tdt=\\\\=-25+25/2=-25/2

The fourth integral. We parametrize \large C_4 as

r(t) = t(0,0)+(1-t)(0,1) = (0, 1-t) for 0≤ t≤ 1 and

r'(t) = (0,-1) so

\large \displaystyle\int_{C_4}xydx+x^2dy=0

So

\large \displaystyle\int_{C}xydx+x^2dy=25-25/2=25/2

Now, let us compute the value using Green's theorem.

According with this theorem

\large \displaystyle\int_{C}Pdx+Qdy=\displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx

where A is the interior of the rectangle.

so A={(x,y) |  0≤ x≤ 5,  0≤ y≤ 1}

We have

\large \displaystyle\frac{\partial Q}{\partial x}=2x\\\\\displaystyle\frac{\partial P}{\partial y}=x

so

\large \displaystyle\iint_{A}(\displaystyle\frac{\partial Q}{\partial x}-\displaystyle\frac{\partial P}{\partial y})dydx=\displaystyle\int_{0}^{5}\displaystyle\int_{0}^{1}xdydx=\displaystyle\int_{0}^{5}xdx\displaystyle\int_{0}^{1}dy=25/2

3 0
3 years ago
Eighteen divided by the sum of two and seven <br> Numerical expression.
Alchen [17]

Answer:

D

Step-by-step explanation:

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