All categories

2 years ago

you recently purchased an item the cost $84.99 before tax. the tax on the item was $7.22. what was the tax rate?

Mathematics

1 answer:

lapo4ka [179]2 years ago

6 0

Is approximately 8.49 percent

You might be interested in

Is the binomial a factor of the polynomial nominal function?

Sedbober [7]

The factors for given equation f(x)= $x^{3} +3x^{2} -25x-75$ .

(x-1) - No
(x-3) - No
(x+3) - Yes
(x-5) - Yes
(x+5) - Yes

Step-by-step explanation:

The given equation is f(x)= $x^{3} +3x^{2} -25x-75$ .

Add 0 at the end of the equation.

$x^{3} +3x^{2} -25x-75$ = 0.

Let us group the given equation,

$(x^{3} +3x^{2}) -(25x-75)$ =0.

⇒ Group 1: $x^{3} +3x^{2}$ .

Group 2: $- 25x +75$ .

Pull out factor from each group,

⇒ Group 1: $(x+3)(x^{2})$ .

Group 2: (x+3) (-25).

Join the two group since both (x+3) is common in both groups.

$(x+3)(x^{2} -25)$ =0.

One of the factor is (x+3).

Other factors are solved by the formula, $a^{2}-b^{2} = (a+b) (a-b)$ .

$(x+3)(x^{2} -25)$ = (x+5) (x-5) .

The other factors are (x+5) and (x-5).

7 0

3 years ago

33.2 x21 area model and standard algorithm

Ad libitum [116K]

Answer:

697.2

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

In circle E, AC=4, CG=6, and AH=3, Find HB and EG. If necessary, round to the tenths place.

bezimeni [28]

Using theorem about secant segments we can write,
AB*AH=AG*AC
AC=4,
CG=6
AG=AC+CG=4+6=10
AH=3
AB= AH+HB=AH+x=3+x
(3+x)*3=10*4
9+3x=40
3x=40-9
3x=31
x=31/3≈10.3
HB≈10.3
EG=HB/2 (as radius and diameter)
EG=10.3/2≈5.2

3 0

3 years ago

What is 502.07 + 1.4? 502.084 502.21 503.47 516.07

solmaris [256]

Answer:

503.47

Step-by-step explanation:

the third one

6 0

3 years ago

Let II be the tangent plane to the graph of f(x, y) = 8 – 2x^2 – 3y^2 at the point (1, 2,-6). Let S, x² + y^2 + z = 4 be another

stealth61 [152]

Let $F(x,y,z)=f(x,y)-z$ . The tangent plane to $f(x,y)$ at (1, 2, -6) has equation

$\nabla F(1,2,-6)\cdot(x-1,y-2,z+6)=0$

We have

$\nabla F(x,y,z)=(-4x,-6y,-1)\implies\nabla F(1,2,-6)=(-4,-12,-1)$

Then the tangent plane has equation

$(-4,-12,-1)\cdot(x-1,y-2,z+6)=0\implies -4(x-1)-12(y-2)-(z+6)=0\implies 4x+12y+z=22$

Let $g(x,y)=4-x^2-y^2$ , and $G(x,y,z)=g(x,y)-z$ . The tangent plane to $S$ at a point $(a,b,c)$ is

$\nabla G(a,b,c)\cdot(x-a,y-b,z-c)=0$

We have

$\nabla G(x,y,z)=(-2x,-2y,-1)\implies \nabla G(a,b,c)=(-2a,-2b,-1)$

so that this plane has equation

$(-2a,-2b,-1)\cdot(x-a,y-b,z-c)=0\implies2ax+2by+z=2a^2+2b^2+c$

In order for this plane to be parallel to the previous plane, we need to have

$\begin{cases}2a=4\\2b=12\end{cases}\implies a=2,b=6\implies g(a,b)=c=-36$

so the point we're looking for is (2, 6, -36).

6 0

3 years ago