All categories

1 year ago

Including a 7% sales tax, an inn charges $109.14 per night. Find the inn's nightly cost before tax is added.

Mathematics

2 answers:

choli [55]1 year ago

8 0

109.14 x 7% = 7.6398
Therefore, 109.14 - 7.6398
= 101.5002

hammer [34]1 year ago

7 0

The answer is $102 per night before tax is added. This is because tax adds up to $7.14.

Hope this helps!

You might be interested in

KHAN ACADEMY PLS HELP

Grace [21]

Answer:

I think its b

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Please help!!!

Dominik [7]

Answer:

Step-by-step explanation:

8+1=9

16+2=18

24+3=27

32+4=36

40+ 5=45

7 0

3 years ago

Please help with this problem:

Olegator [25]

He added 8 to both sides. He did the write thing.

6 0

2 years ago

2(x+x+12) > 100 (why does the 20 character thing exist)

arlik [135]

Answer:

x > 19

Step-by-step explanation:

2(x+x+12) > 100

2(2x+12) > 100

4x+24 > 100

4x > 76

x > 19

3 0

3 years ago

find the coordinates of the point P on the parabola y=1-x^2 with domain 0≤x≤1 that minimize the area of the triangle enclosed by

coldgirl [10]

Let point P be with coordinates $(x_0,y_0).$ Find the equation of the tangent line.

1. If $y=1-x^2,$ then $y'=-2x.$

2. The equation of the tangent line at point P is

$y-y_0=-2x_0(x-x_0).$

Find x-intercept and y-intercept of this line:

when x=0, then $y=y_0+2x_0^2;$
when y=0, then $x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.$

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

$A=\dfrac{1}{2}\cdot (2x_0^2+y_0)\cdot \left(\dfrac{y_0+2x_0^2}{2x_0}\right)=\dfrac{(y_0+2x_0^2)^2}{4x_0}.$

Since point P is on the parabola, then $y_0=1-x_0^2$ and

$A=\dfrac{(1-x_0^2+2x_0^2)^2}{4x_0}=\dfrac{(1+x_0^2)^2}{4x_0}.$

Find the derivative A':

$A'=\dfrac{2(1+x_0^2)\cdot 2x_0\cdot 4x_0-4(1+x_0^2)^2}{16x_0^2}=\dfrac{12x_0^4+8x_0^2-4}{16x_0^2}.$

Equate this derivative to 0, then

$12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.$

And

$y_0=1-\left(\pm\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{2}{3}.$

Answer: two points: $P_1\left(-\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right), P_2\left(\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right).$

6 0

3 years ago