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yKpoI14uk [10]
3 years ago
6

An item is regularly priced at $95 . it is on sale for 40% off the regular price. how much (in dollars) is discounted from the r

egular price?
Mathematics
1 answer:
sergiy2304 [10]3 years ago
8 0
95.00*40%=
Answer=$38.00
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3x-2(x+3)=4x-9-x is the question
kozerog [31]

Answer:

\large\boxed{x = 7}

Step-by-step explanation:

3x - 2(x + 3) = 4x - 9 - x

Start by multiplying 2 by the values in the parenthesis

3x - 2x + 6 = 4x - 9 - x

Subtract 3x - 2x and 4x - x

x + 5 = 3x - 9

Add 9 to both sides of the equation

x + 14 = 3x

Subtract x from both sides of the equation

14 = 2x

Divide both sides of the equation by 2

\large\boxed{x = 7}

Hope this helps :)

8 0
3 years ago
Your answer should be a polynomial in standard form. ( − 8 p 3 + 9 p 2 q − 5 p q 2 ) + ( 8 p 3 − p q + 5 q 2 ) = (−8p ​3 ​​ +9p
gavmur [86]
9p^2-6p5q^2 in standard form it would be: 9p^2+5q^2-6p
5 0
3 years ago
Find the measure of angle A
AVprozaik [17]

Answer: The answer is 35 degrees

5 0
3 years ago
Describe how to transform:
Anna71 [15]

Answer:

x^\frac{35}{6}

Step-by-step explanation:

The expression to transform is:

(\sqrt[6]{x^5})^7

Let's work first on the inside of the parenthesis.

Recall that the n-root of an expression can be written as a fractional exponent of the expression as follows:

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Therefore \sqrt[6]{a} = a^{\frac{1}{6}}

Now let's replace a with x^{5} which is the algebraic form we are given inside the 6th root:

\sqrt[6]{x^5} = (x^5)^{\frac{1}{6}}

Now use the property that tells us how to proceed when we have  "exponent of an exponent":

(a^n)^m= a^{n*m}

Therefore we get:  (x^5)^{\frac{1}{6}}=x^{\frac{5}{6}}}

Finally remember that this expression was raised to the power 7, therefore:

[tex](\sqrt[6]{x^5})^7=(x^\frac{5}{6})^7=x^\frac{35}{6}[/tex]

An use again the property for the exponent of a exponent:

8 0
2 years ago
4x + 3 < 3 OR 4x + 3 > 19
JulijaS [17]

4x + 3 > 19

Step-by-step explanation:

4x + 3 > 19  \\ 4x + 3 - 3 > 19 - 3 \\ 4x > 16 \\ 4x \div 4 > 16 \div 4 \\ x > 4

4 0
2 years ago
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