store owner decides to mark up all items by 30%. What is the selling price of an item that originally cost $10?

Mathematics

1 answer:

Artist 52 [7]2 years ago

6 0

0.030 × 10 =.0.3. 0.3 +10 = 10.3

You might be interested in

Und your answers to the near<br> 6(7y + 8) = 22

Vladimir [108]

Answer:

y=⁻ $\\\frac{13}{21}$ should be your answer in decimal form its y=-0.619047

Step-by-step explanation:

4 0

2 years ago

What is the quadratic regression equation?

aniked [119]

Answer:

y=-0.078x^2+8.407x+-92.892

Step-by-step explanation:

but your x and y in desmo than go under the table do y1=ax^2 then under the 2 put a 1 than +bx1+c than you round your a b and c to the nearest thousandth than after that add a b and c then after you get your answer than but x^2 behind a than x behind b then nothing behind c hope this works for you

7 0

2 years ago

Type the correct answer in each box. Use numerals instead of words,

xeze [42]

Answer:

x = -6,3,15,-12; y = 11,5,-3,15

Step-by-step explanation:

Domain is the x value, so plugged in the x values into the equation and got the y values or range.

8 0

2 years ago

Your Assignment What is the fractional equivalent of the repeating decimal n = 0.1515... ? Answer the questions to find out.

Savatey [412]

Answer:

The fractional equivalent of the repeating decimal would be 15/99 and 0.1515... is a repeating decimal so it doesn't end, but in the shown number is 4 digits.

3 0

2 years ago

Find the x-values that are solutions of the equation 5cot^2x-15=0

julia-pushkina [17]

First solve for the trig function 'cot'
$cot^2 x = \frac{15}{5} = 3$

Next take the sqrt of both sides (include plus/minus)
$cot x = \pm \sqrt{3}$

Now take reciprocal of both sides, this will change trig function to 'tan'
(cot = 1/tan)

$tan x = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}$

Finally use the unit circle or inverse tan on your calculator to find x.
There will be 4 solutions, one for each quadrant.

$x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$

3 0

2 years ago