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gladu [14]
3 years ago
7

You are comparing prices from 2 office supply stores. Your office needs 5 cases of blue paper. Home & Office Headquarters li

sts a case of paper at $25.85 with a 10% discount on an order of 5 cases or more. Office Supplies R Us lists a case of paper at $27,36 with a 15% discount on 5 cases or more. Delivery costs from Home & Office Headquarters are $2.50 per case. Office Supplies R Us will deliver for $10 an order. What is the least amount that you would have to spend for the paper?
Mathematics
1 answer:
melomori [17]3 years ago
8 0


5x[$25.85-($25.85 x 0.10) ] + (5 x 2.50)=$128.83 compared to[ 5 x 27.36( $27.36 x 0.15)]+ 10.00= 126.28 is the least amount.

answer is ------> $126.28

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What is the complete factorization of the polynomial below
zepelin [54]

Answer:

B. (x+4)(x+i)(x-i)

Step-by-step explanation:

Let P(x)=x^3-4x^2+x-4

We can factor this polynomial by grouping:

P(x)=x^2(x-4)+1(x-4)

We factor further to obtain:

P(x)=(x^2+1)(x-4)

P(x)=(x^2-\sqrt{-1}^2)(x-4)

We apply difference of two squares to get:

P(x)=(x-i)(x+i)(x-4)

7 0
3 years ago
Read 2 more answers
2x(9-5x) - (-4x-36x)
allochka39001 [22]
I got 48x
First by distributing 2x to 9-5x=
18x-10x
Then adding a 1 in front of - making -4x-36x to 4x+36x=40x
8x+40x=48x
7 0
3 years ago
Como resolver y=1-2.5
oksian1 [2.3K]

I don't really speak Spanish but I'll guess that says

 How do I solve  y = 1 - 2.5

Unless I'm missing something, there's not even any algebra here.  

The way we subtract a bigger number from a smaller number is first we subtract the smaller number from the bigger number the regular way, then we prepend a minus sign.


y = 1 - 2.5 = -(2.5 - 1) = -1.5



7 0
4 years ago
Hi can you help me in mate plis simplify the following algebraic expressions​
mars1129 [50]

Answer:

\frac{3 {x}^{2} }{ {y}^{2} {z}^{6}  }

Step-by-step explanation:

I have attached the explanation above. hopefully this will help

4 0
3 years ago
Find the derivative.
krek1111 [17]

Answer:

\displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}

General Formulas and Concepts:

<u>Algebra I</u>

Terms/Coefficients

  • Expanding/Factoring

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:                                                                           \displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

\displaystyle f(x) = \frac{\sqrt{x}}{e^x}

<u>Step 2: Differentiate</u>

  1. Derivative Rule [Quotient Rule]:                                                                   \displaystyle f'(x) = \frac{(\sqrt{x})'e^x - \sqrt{x}(e^x)'}{(e^x)^2}
  2. Basic Power Rule:                                                                                         \displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}(e^x)'}{(e^x)^2}
  3. Exponential Differentiation:                                                                         \displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{(e^x)^2}
  4. Simplify:                                                                                                         \displaystyle f'(x) = \frac{\frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x}{e^{2x}}
  5. Rewrite:                                                                                                         \displaystyle f'(x) = \bigg( \frac{e^x}{2\sqrt{x}} - \sqrt{x}e^x \bigg) e^{-2x}
  6. Factor:                                                                                                           \displaystyle f'(x) = \bigg( \frac{1}{2\sqrt{x}} - \sqrt{x} \bigg)e^\big{-x}

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

7 0
3 years ago
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