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Wewaii [24]
3 years ago
11

If a wholesale price of an item is $250 and the markup is 25%. how much will you pay for the item

Mathematics
1 answer:
Eddi Din [679]3 years ago
4 0
$250 x 0.25 = 62.5
62.5 + 250 == $312.50
you will pay $312.50
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Use this graph to answer the following two questions.
umka21 [38]
This is quite interesting let me know if someone responds…
8 0
2 years ago
Read 2 more answers
Find the distance between points P(8, 2) and Q(3, 8) to the nearest tenth
olga nikolaevna [1]
(11/2 , 5) sorry if im wrong im not too sure
4 0
3 years ago
X - 3y +3=0
Arte-miy333 [17]

Answer:

We know that for a line:

y = a*x + b

where a is the slope and b is the y-intercept.

Any line with a slope equal to -(1/a) will be perpendicular to the one above.

So here we start with the line:

3x + 4y + 5 = 0

let's rewrite this as:

4y = -3x - 5

y = -(3/4)*x - (5/4)

So a line perpendicular to this one, has a slope equal to:

- (-4/3) = (4/3)

So the perpendicular line will be something like:

y = (4/3)*x + c

We know that this line passes through the point (a, 3)

this means that, when x = a, y must be equal to 3.

Replacing these in the above line equation, we get:

3 = (4/3)*a + c

c = 3 - (4/3)*a

Then the equation for our line is:

y = (4/3)*x + 3 - (4/3)*a

We can rewrite this as:

y = (4/3)*(x -a) + 3

now we need to find the point where this line ( y = -(3/4)*x - (5/4)) and the original line intersect.

We can find this by solving:

(4/3)*(x -a) + 3 =  y = -(3/4)*x - (5/4)

(4/3)*(x -a) + 3  = -(3/4)*x - (5/4)

(4/3)*x - (3/4)*x = -(4/3)*a - 3 - (5/4)

(16/12)*x - (9/12)*x = -(4/3)*a - 12/4 - 5/4

(7/12)*x = -(4/13)*a - 17/4

x = (-(4/13)*a - 17/4)*(12/7) = - (48/91)*a - 51/7

And the y-value is given by inputin this in any of the two lines, for example with the first one we get:

y =  -(3/4)*(- (48/91)*a - 51/7) - (5/4)

  = (36/91)*a + (153/28) - 5/4

Then the intersection point is:

( - (48/91)*a - 51/7,  (36/91)*a + (153/28) - 5/4)

And we want that the distance between this point, and our original point (3, a) to be equal to 4.

Remember that the distance between two points (a, b) and (c, d) is:

distance = √( (a - c)^2 + (b - d)^2)

So here, the distance between (a, 3) and ( - (48/91)*a - 51/7,  (36/91)*a + (153/28) - 5/4) is 4

4 = √( (a + (48/91)*a + 51/7)^2 + (3 -  (36/91)*a + (153/28) - 5/4 )^2)

If we square both sides, we get:

4^2 = 16 =  (a + (48/91)*a + 51/7)^2 + (3 -  (36/91)*a - (153/28) + 5/4 )^2)

Now we need to solve this for a.

16 = (a*(1 + 48/91)  + 51/7)^2 + ( -(36/91)*a  + 3 - 5/4 + (153/28) )^2

16 = ( a*(139/91) + 51/7)^2 + ( -(36/91)*a  - (43/28) )^2

16 = a^2*(139/91)^2 + 2*a*(139/91)*51/7 + (51/7)^2 +  a^2*(36/91)^2 + 2*(36/91)*a*(43/28) + (43/28)^2

16 = a^2*( (139/91)^2 + (36/91)^2) + a*( 2*(139/91)*51/7 + 2*(36/91)*(43/28)) +  (51/7)^2 + (43/28)^2

At this point we can see that this is really messy, so let's start solving these fractions.

16 = (2.49)*a^2 + a*(23.47) + 55.44

0 = (2.49)*a^2 + a*(23.47) + 55.44 - 16

0 = (2.49)*a^2 + a*(23.47) + 39.44

Now we can use the Bhaskara's formula for quadratic equations, the two solutions will be:

a = \frac{-23.47  \pm  \sqrt{23.47^2 - 4*2.49*39.4}  }{2*2.49} \\\\a =  \frac{-23.47  \pm  12.57 }{4.98}

Then the two possible values of a are:

a = (-23.47 + 12.57)/4.98  = -2.19

a = (-23.47 - 12.57)/4.98 = -7.23

4 0
3 years ago
A survey of 500 high school students was taken to determine their favorite chocolate candy. Of the 500 students surveyed, 64 lik
Lyrx [107]

Answer:

466 student

Step-by-step explanation:

The number of students that like at most 2 kinds of these chocolate candies can be gotten from subtracting the number of students that like more than 2 candies from the total number of high school students.

Since 34 like all three kinds of chocolate candy, hence:

Number of students that like at most 2 kinds of these chocolate candies = 500 - 34 = 466

466 students like at most 2 kinds of these chocolate candies

6 0
3 years ago
Find the tenth term of the geometric sequence, given the first term and common ratio.
Natalija [7]

Answer:

a_{10}=\dfrac{1}{128}

Step-by-step explanation:

In the geometric series:

a_1=4\\ \\r=\dfrac{1}{2}

The nth term of the geometric sequence can be calculated using formula

a_n=a_1\cdot r^{n-1}

In your case, n = 10, then

a_{10}\\ \\=4\cdot \left(\dfrac{1}{2}\right)^{10-1}\\ \\=4\cdot \left(\dfrac{1}{2}\right)^9\\ \\=2^2\cdot \dfrac{1}{2^9}\\ \\=\dfrac{1}{2^{9-2}}\\ \\=\dfrac{1}{2^7}\\ \\=\dfrac{1}{128}

8 0
3 years ago
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