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Arte-miy333 [17]
3 years ago
11

alana and petra go to a used book sale, where every book is the same price. Alana buys 5 books and Petra buys 4 books. They run

into their friend Angus, who really loves to read, and discover that he has bought 19 books. Angus received a $10 discount, but he still spent $5 more than Alana and Petra combined. How much does one book cost?
Mathematics
1 answer:
Pachacha [2.7K]3 years ago
8 0
After understanding the problem, formulate equations to determine the unknown. Let x be the price of one book. Equating Alana&Petra's expense with Angus' would be

5x + 4x +5 = 19x - 10

The 5 in the left side is needed because Angus' expense is $5 more. Also, the 10 on the right side represents the $10 discount. Solving the equation,

19x = 9x = 5+10
10x = 15
x = $1.5

Therefore, the book costs $1.5. 
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-43 – 9x = -7 – 3(5x + 6)<br> Answer?
lilavasa [31]

Answer:

Simplifying

9x + 7 = 5x + -3

Reorder the terms:

7 + 9x = 5x + -3

Reorder the terms:

7 + 9x = -3 + 5x

Solving

7 + 9x = -3 + 5x

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-5x' to each side of the equation.

7 + 9x + -5x = -3 + 5x + -5x

Combine like terms: 9x + -5x = 4x

7 + 4x = -3 + 5x + -5x

Combine like terms: 5x + -5x = 0

7 + 4x = -3 + 0

7 + 4x = -3

Add '-7' to each side of the equation.

7 + -7 + 4x = -3 + -7

Combine like terms: 7 + -7 = 0

0 + 4x = -3 + -7

4x = -3 + -7

Combine like terms: -3 + -7 = -10

4x = -10

Divide each side by '4'.

x = -2.5

Simplifying

x = -2.5

make me brainiest plz!!!

Step-by-step explanation:

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2 years ago
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What is the exact situation
Luda [366]
Number Two or B would be correct
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2 years ago
Analyze the expression and then choose the sentences describing whether the binomial theorem can be used to factor it. Check all
daser333 [38]

Answer:

Step-by-step explanation:

N is 3 bc #of terms minus 1

Exponents go from 3>2>x>none

Can be factored using binomial theorem

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If <img src="https://tex.z-dn.net/?f=%5Cmathrm%20%7By%20%3D%20%28x%20%2B%20%5Csqrt%7B1%2Bx%5E%7B2%7D%7D%29%5E%7Bm%7D%7D" id="Tex
Harman [31]

Answer:

See below for proof.

Step-by-step explanation:

<u>Given</u>:

y=\left(x+\sqrt{1+x^2}\right)^m

<u>First derivative</u>

\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If  $f(g(x))$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=f'(g(x))\:g'(x)$\\\end{minipage}}

<u />

<u />\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If  $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}

<u />

\begin{aligned} y_1=\dfrac{\text{d}y}{\text{d}x} & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{2x}{2\sqrt{1+x^2}} \right)\\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(1+\dfrac{x}{\sqrt{1+x^2}} \right) \\\\ & =m\left(x+\sqrt{1+x^2}\right)^{m-1} \cdot \left(\dfrac{x+\sqrt{1+x^2}}{\sqrt{1+x^2}} \right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^{m-1}  \cdot \left(x+\sqrt{1+x^2}\right)\\\\ & = \dfrac{m}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m\end{aligned}

<u>Second derivative</u>

<u />

\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If  $y=uv$  then:\\\\$\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}

\textsf{Let }u=\dfrac{m}{\sqrt{1+x^2}}

\implies \dfrac{\text{d}u}{\text{d}x}=-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}

\textsf{Let }v=\left(x+\sqrt{1+x^2}\right)^m

\implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{m}{\sqrt{1+x^2}} \cdot \left(x+\sqrt{1+x^2}\right)^m

\begin{aligned}y_2=\dfrac{\text{d}^2y}{\text{d}x^2}&=\dfrac{m}{\sqrt{1+x^2}}\cdot\dfrac{m}{\sqrt{1+x^2}}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)^\frac{3}{2}}\\\\&=\dfrac{m^2}{1+x^2}\cdot\left(x+\sqrt{1+x^2}\right)^m+\left(x+\sqrt{1+x^2}\right)^m\cdot-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\\\\ &=\left(x+\sqrt{1+x^2}\right)^m\left(\dfrac{m^2}{1+x^2}-\dfrac{mx}{\left(1+x^2\right)\sqrt{1+x^2}}\right)\\\\\end{aligned}

              = \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\right)\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)

<u>Proof</u>

  (x^2+1)y_2+xy_1-m^2y

= (x^2+1) \dfrac{\left(x+\sqrt{1+x^2}\right)^m}{1+x^2}\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m

= \left(x+\sqrt{1+x^2}\right)^m\left(m^2-\dfrac{mx}{\sqrt{1+x^2}}\right)+\dfrac{mx}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^m-m^2\left(x+\sqrt{1+x^2\right)^m

= \left(x+\sqrt{1+x^2}\right)^m\left[m^2-\dfrac{mx}{\sqrt{1+x^2}}+\dfrac{mx}{\sqrt{1+x^2}}-m^2\right]

= \left(x+\sqrt{1+x^2}\right)^m\left[0]

= 0

8 0
1 year ago
16x+5=12x-6, 4x+5=-6 what property is that
Nuetrik [128]

Answer:

subtraction property

Step-by-step explanation:

because you subtracted 12x from 16x to get 4x+5=-6

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3 years ago
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