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3 years ago

What is the value of X??

Mathematics

2 answers:

mars1129 [50]3 years ago

5 0

Answer:

Step-by-step explanation:

amid [387]3 years ago

5 0

Answer:

Step-by-step explanation:

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Please explain what mark up is to me. Keep it simple

iogann1982 [59]

Hey there!

Markup is the percentage that the cost of an item is increased.

For example, if a company bought a pair of headphones that costed $100 and they sold it for $150, the markup was 50%, since half of the original cost was added. if the markup were 100% it would have been sold for $200, since it added 100% of the original.

I hope this helps!

8 0

3 years ago

Use the Distributive Property to simplify the expression 6 • 68.

Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

6*68=408

------------------

48/10=4.8

----------------

3.8=x/7.7

x=3.8*7.7x

x=29.26

---------------

x+5.1=13.4

x=13.4-5.1

x=8.3

---------------

8x=64

x=64/8

x=8

--------------

22=m+1

m=22-1

m=21

----------

45-4x for x=5

45-4(5)=45-20=25

---------------------------

256, 64, 16, 4

7 0

3 years ago

Two rectangles have the same area of 70m2. It is given that the length of the first rectangle is 4m longer and the width is 2m s

NeX [460]

Answer:

second rectangle

longer = 10

width = 7

first rectangle

longer = 14

width= 5

Step-by-step explanation:

7 0

3 years ago

6-2x=y 14+3x=y plzzz helllllllllllllllp

pashok25 [27]

Answer:

(- $\frac{8}{5}$ , $\frac{46}{5}$ )

Step-by-step explanation:

Given the2 equations

6 - 2x = y → (1)

14 + 3x = y → (2)

Substitute y = 14 + 3x into (1)

6 - 2x = 14 + 3x ( subtract 3x from both sides )

6 - 5x = 14 ( subtract 6 from both sides )

- 5x = 8 ( divide both sides by - 5 )

x = - $\frac{8}{5}$

Substitute x = - $\frac{8}{5}$ into either of the 2 equations for the corresponding value of y

Substituting into (1)

y = 6 - 2(- $\frac{8}{5}$ ) = $\frac{30}{5}$ + $\frac{16}{5}$ = $\frac{46}{5}$

solution is ( - $\frac{8}{5}$ , $\frac{46}{5}$ )

8 0

3 years ago

A triangular lamina has vertices (0, 0), (0, 1) and (c, 0) for some positive constant c. Assuming constant mass density, show th

a_sh-v [17]

The equation of the line through (0, 1) and (c, 0) is

y - 0 = (0 - 1)/(c - 0) (x - c) ==> y = 1 - x/c

Let L denote the given lamina,

L = {(x, y) : 0 ≤ x ≤ c and 0 ≤ y ≤ 1 - x/c}

Then the center of mass of L is the point $(\bar x,\bar y)$ with coordinates given by

$\bar x = \dfrac{M_x}m \text{ and } \bar y = \dfrac{M_y}m$

where $M_x$ is the first moment of L about the x-axis, $M_y$ is the first moment about the y-axis, and m is the mass of L. We only care about the y-coordinate, of course.

Let ρ be the mass density of L. Then L has a mass of

$\displaystyle m = \iint_L \rho \,\mathrm dA = \rho\int_0^c\int_0^{1-\frac xc}\mathrm dy\,\mathrm dx = \frac{\rho c}2$

Now we compute the first moment about the y-axis:

$\displaystyle M_y = \iint_L x\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}x\,\mathrm dy\,\mathrm dx = \frac{\rho c^2}6$

Then

$\bar y = \dfrac{M_y}m = \dfrac{\dfrac{\rho c^2}6}{\dfrac{\rho c}2} = \dfrac c3$

but this clearly isn't independent of c ...

Maybe the x-coordinate was intended? Because we would have had

$\displaystyle M_x = \iint_L y\rho\,\mathrm dA = \rho \int_0^c\int_0^{1-\frac xc}y\,\mathrm dy\,\mathrm dx = \frac{\rho c}6$

and we get

$\bar x = \dfrac{M_x}m = \dfrac{\dfrac{\rho c}6}{\dfrac{\rho c}2} = \dfrac13$

8 0

3 years ago