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frosja888 [35]
2 years ago
6

Caitlin makes candles and sells them in her online shop. Tall candles cost $8 each and short candles cost $5 each. last month, s

he sold a total of 91 candles and made $572.
how many tall candles did she sell?

how many short candles did she sell?
Mathematics
1 answer:
snow_tiger [21]2 years ago
3 0

Answer:

tall candles = 39

Short candles = 52

Step-by-step explanation:

let tall candles be x

and short candles be y

No. Of candles sold = x + y =91

Money made from candles will be equal to = 8x + 5y = $572

These both are simultaneous linear equations.

x + y =91 ( multiply by 5)

5x + 5y = 455

8x + 5y = $572

Subtract them:

-3x = -117

x = -117/3

x = 39

put value of x in

x + y = 91

39 + y = 91

y = 91 - 39

y = 52

x shows tall candles so they are 39

y shows short candles so they are 52

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Anon25 [30]

Answer:

The required equation of the graph with the coordinates of (4, -1) will be:

  • y=\left(x-4\right)^2-1

The graph is also attached.

Step-by-step explanation:

Given the coordinates (4, -1)

If we put the coordinate values (4, -1) in the given equations, we determine that only y=\left(x-4\right)^2-1 is the valid equation as it satisfies the given coordinate value.

For example, putting  (4, -1) in the equation

y=\left(x-4\right)^2-1

\left(-1\right)=\left(4-4\right)^2-1

-1=\left(0\right)^2-1

-1=0-1

-1=-1

So, the equation y=\left(x-4\right)^2-1 is true for the coordinate values (4, -1).

Therefore, the required equation of the graph with the coordinates of (4, -1) will be:

  • y=\left(x-4\right)^2-1

Please check, the graph is also attached.

3 0
2 years ago
A point P is moving along the curve whose equation is y = \sqrt x . Suppose that x is increasing at the rate of 4 units/s when x
Mice21 [21]

Answer:3 units/s

Step-by-step explanation:

Given

y=\sqrt{x}

Point P lie on this curve so any general point on curve can be written as (x,\sqrt{x})

and \frac{\mathrm{d} x}{\mathrm{d} t}=4 units/s

Distance between Point P and (2,0)

P=\sqrt{(x-2)^2+(\sqrt{x}-0)^2}

P at x=3 P=2

rate at which distance is changing is

\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{\mathrm{d} \sqrt{(x-2)^2+(\sqrt{x}-0)^2}}{\mathrm{d} t}

\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2x-3}{\sqrt{(x-2)^2+(\sqrt{x}-0)^2}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} P}{\mathrm{d} t}=\frac{2\times 3-3}{2\times 2}\times 4=3 units/s

8 0
3 years ago
Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (4, −5, 2) and parallel
Nataliya [291]

Answer:

Step-by-step explanation:

From the given information, the symmetric equations for the line pass through(4, -5, 2) i.e (x_o, y_o, z_o) and are parallel to \dfrac{x+5}{1} = \dfrac{y}{2}= \dfrac{z-3}{1}

The parallel vector to the line i + zj+k = ai + bj + ck

Hence, the equation for the line is :

x = x_o + at \\ \\ x = y_o + bt \\ \\ x = z_o + ct

x = 4 + t

y = -5 + 2t

z = 2 + t

Thus, x, y, z = ( 4+t, -5+2t, 2+t )

The symmetric equation can now be as follows:

\begin  {vmatrix} x = 4+ t   \\ \\  \dfrac{x-4}{1} = t  \begin {vmatirx} \end {vmatrix}\begin {vmatrix} y = - 5+2t  \\ \\ \dfrac{y+5}{2}  =t      \end {vmatrix}\begin {vmatrix} z =2+t  \\ \\ \dfrac{z-2}{1}  =t      \end {vmatrix}

∴

\dfrac{x-4}{1}= \dfrac{y+5}{2}=\dfrac{z-2}{1}

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astraxan [27]
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No solutions
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2 years ago
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meriva
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