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2 years ago

What is the discount and sale price of a $300 item that has been discounted 10%?

Mathematics

2 answers:

faust18 [17]2 years ago

7 0

Answer:

The discount is $30, making the sale price $270.

Step-by-step explanation:

maxonik [38]2 years ago

5 0

Answer: The discount is $30, making the sale price $270.

Step-by-step explanation: Take your initial value times the percent of your discount to find the discount amount: in this case, it is 300 * .10 = 30. Then you can subtract that value from your initial value, 300-30=270 to get the sale price.

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If T(n) = 7- n , what is the 2nd term?

Gelneren [198K]

$T(2)=7-2=5$

7 0

3 years ago

Use the Distance Formula to find the distance between D(2, 0) and E(8, 6).

Mazyrski [523]

Answer:

The answer is

<h2> $6 \sqrt{2} \: \: \: or \: \: \: 8.50 \: \: \: \: units$ </h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question

The points are D(2, 0) and E(8, 6

The distance between them is

$|DE| = \sqrt{ ({2 - 8})^{2} + ({0 - 6})^{2} } \\ = \sqrt{( { - 6})^{2} + ( { - 6})^{2} } \\ = \sqrt{36 + 36} \\ = \sqrt{72} \: \: \: \: \: \: \: \: \: \: \\ = 6 \sqrt{2} \: \: \: \: \: \: \: \: \: \: \\ = 8.4852813$

We have the final answer as

$6 \sqrt{2} \: \: \: or \: \: \: 8.50 \: \: \: \: units$

Hope this helps you

5 0

3 years ago

NEED ANSWER ASAP!!!! <br> y=−x^2−2x+3 <br> When is g(x) = 0?

Crazy boy [7]

Answer:

$y=3$

Step-by-step explanation:

Since g(x)=0, you input 0 into every x

$y=-(0)^{2} -2(0)+3$

Which then becomes:

$y=0+3$

Take out the 0 and there's your answer

7 0

2 years ago

GUYSSSS PLEASEEEEEEEEEE HELP!!!!!! Your friend is using Descartes’ Rule of Signs to find the number of negative real roots of <i

jeka57 [31]

Answer:

your answer should be a

Step-by-step explanation:

7 0

2 years ago

Solve the following by using mathematical Induction. For >/ 1

cluponka [151]

Answer:

See explanation

Step-by-step explanation:

Prove that

$1^2+2^2+3^3+...+n^2=\dfrac{1}{6}n(n+1)(2n+1)$

1. When $n=1,$ we have

in left part $1^2=1;$

in right part $\dfrac{1}{6}\cdot 1\cdot (1+1)\cdot (2\cdot 1+1)=\dfrac{1}{6}\cdot 1\cdot 2\cdot 3=1.$

2. Assume that for all $k$ following equality is true

$1^2+2^2+3^3+...+k^2=\dfrac{1}{6}k(k+1)(2k+1)$

3. Prove that for $k+1$ the following equality is true too.

$1^2+2^2+3^3+...+(k+1)^2=\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)$

Consider left part:

$1^2+2^2+3^2+...+(k+1)^2=\\ \\=(1^2+2^2+3^3+...+k^2)+(k+1)^2=\\ \\=\dfrac{1}{6}k(k+1)(2k+1)+(k+1)^2=\\ \\=(k+1)\left(\dfrac{1}{6}k(2k+1)+k+1\right)=\\ \\=(k+1)\dfrac{2k^2+k+6k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+7k+6}{6}=\\ \\=(k+1)\dfrac{2k^2+4k+3k+6}{6}=\\ \\=(k+1)\dfrac{2k(k+2)+3(k+2)}{6}=\\ \\=(k+1)\dfrac{(k+2)(2k+3)}{6}$

Consider right part:

$\dfrac{1}{6}(k+1)((k+1)+1)(2(k+1)+1)=\\ \\\dfrac{1}{6}(k+1)(k+2)(2k+3)$

We get the same left and right parts, so the equality is true for $k+1.$

By mathematical induction, this equality is true for all n.

3 0

3 years ago