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3 years ago

10

Terry bought a sweater for$24 before tax he had a coupon for 1/4 off the sales tax rate was 6%. How much did terry pay for the s

weater, including tax?

1 answer:

Ira Lisetskai [31]3 years ago

8 0

Answer: Terry paid $18.36 for the sweater.

Step-by-step explanation:

Given: Sales price of sweater = $24

Coupon states $\dfrac14$ off the sales, i.e. Terry will get $\dfrac14\times24=\$6$ discount.

Also, tax rate = 6%

Tax will be charged = 6% of Sales price of sweater

= 0.06 x 6

= $0.36

Now , Final amount for Terry = Sales price - discount +tax

= $(24-6+0.36)= $18.36

Hence, Terry paid $18.36 for the sweater.

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a baseball helmet cost $42 coach Rodger said he would pay $274 for 7 helmets is the exact answer reasonable

Zepler [3.9K]

No cause if I helmet cost $42 , 7 helmets will be $294.
$42 = 1 helmet
$? = 7 helmets
cross multiply = 42 × 7 = $294

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4 years ago

Picasso's Phone Shop charges $65.54 for its activation fee plus $13.10 for every gigabyte, g, of data you use over your limit. T

konstantin123 [22]

Answer:

$13.10 represents the rate of change

Step-by-step explanation:

13.1 is the only value in the equation with a variable (g). There is an additional $13.10 charge for every gigabyte used over the limit, and this is the rate of change.

For example, if 5 gigabytes were used over the limit, the charge would be 13.1(5) or $65.50

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2 years ago

If -2a > 6, then _____.<br><br> a > -3<br> a > 3<br> a < 3<br> a < -3

shutvik [7]

Answer:

a<-3

bc -3(-2)=6

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Step-by-step explanation:

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3 years ago

Find an equation of the sphere that passes through the origin and whose center is (-2, 2, 3). Be sure that your formula is monic

Andrei [34K]

Answer:

$\bold{(x+2))^2+(y-2)^2+(z-2)^2=17}$

Step-by-step explanation:

Given the center of sphere is: (-2, 2, 3)

Passes through the origin i.e. (0, 0, 0)

To find:

The equation of the sphere ?

Solution:

First of all, let us have a look at the equation of a sphere:

$(x-a)^2+(y-b)^2+(z-c)^2=r^2$

Where ( $x,y,z$ ) are the points on sphere.

$(a, b, c)$ is the center of the sphere and

$r$ is the radius of the sphere.

Radius of the sphere is nothing but the distance between any point on the sphere and the center.

We are given both the points, so we can use distance formula to find the radius of the given sphere:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

Here,

$x_1 =0 \\y_1 =0 \\z_1 =0 \\x_2 =-2 \\y_2 =2 \\z_2 =3$

So, Radius is:

$r = \sqrt{(-2-0)^2+(2-0)^2+(3-0)^2}\\\Rightarrow r = \sqrt{4+4+9} = \sqrt{17}$

Therefore the equation of the sphere is:

$(x-(-2))^2+(y-2)^2+(z-2)^2=(\sqrt{17})^2\\\bold{(x+2))^2+(y-2)^2+(z-2)^2=17}$

4 0

4 years ago

which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater

Nutka1998 [239]

Answer:

$\frac{\sqrt[3]{16y^4}}{x^2}$

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices $\sqrt[n]{a} = a^{\frac{1}{n}}$

So,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ gives

$\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}$

Also from laws of indices

$(ab)^n = a^nb^n$

So, the above expression can be further simplified to

$\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}$

Multiply the exponents gives

$\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

Substitute $2^5$ for 32

$\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$

From laws of indices

$\frac{a^m}{a^n} = a^{m-n}$

This law can be applied to the expression above;

$\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}$ becomes

$2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}$

Solve exponents

$2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}$

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$

From laws of indices,

$a^{-n} = \frac{1}{a^n}$ ; So,

$2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}$ gives

$\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}$

The expression at the numerator can be combined to give

$\frac{(2y)^{\frac{4}{3}}}{x^2}$

Lastly, From laws of indices,

$a^{\frac{m}{n} = \sqrt[n]{a^m}$ ; So,

$\frac{(2y)^{\frac{4}{3}}}{x^2}$ becomes

$\frac{\sqrt[3]{(2y)}^{4}}{x^2}$

$\frac{\sqrt[3]{16y^4}}{x^2}$

Hence,

$\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }$ is equivalent to $\frac{\sqrt[3]{16y^4}}{x^2}$

8 0

3 years ago