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3 years ago

5

1. Your bank account has -$15 in it. You deposit $5 per day. (a) How much money is in your account after 5 days? Show your work.

(b) After you have made all of your deposits, you withdraw $2 a day until your account balance is $0. How many days will you withdraw $2? Show your work.PLEZ HELP!

2 answers:

ki77a [65]3 years ago

6 0

A) 15 + 5 X 5 = 15 + 25 = 40

B) 40/2 = 20

you can pull out $2 everyday for 20 days until you have no money left

Harman [31]3 years ago

4 0

A) 5 x 5 = 25$
25 - 15 = 10$
Answer: 10$

b) 10/2= 5
Answer: 5 days

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find the equation of the pair of lines perpendicular to the lines pair represented by the equation ax^2-2hxy+by^2=0 and passing

Naddika [18.5K]

The equation of the pair of lines perpendicular to the lines given equation is $b x^{2}-2 h x y+a y^{2}=0$ .

Solution:

Given equation is $a x^{2}+2 h x y+b y^{2}=0$ .

Let $m_1$ and $m_2$ be the slopes of the given lines.

Sum of the roots = $-\frac{\text {coefficient of } x y}{\text {coefficient of } y^{2}}$

$$m_1+m_2=\frac{-2h}{b}$ – – – – – (1)

Product of the roots = $-\frac{\text {coefficient of } x^2}{\text {coefficient of } y^{2}}$

$$m_1 \cdot m_2=\frac{a}{b}$ – – – – – (2)

The required lines are perpendicular to these lines.

Slopes of the required lines are $$-\frac{1}{m_{1}} \text { and }-\frac{1}{m_{2}}$

Required lines also passes through the origin,

therefore their y-intercepts are 0.

Hence their equations are:

$$y=-\frac{1}{m_{1}}x \text { and }y=-\frac{1}{m_{2}}x$

Do cross multiplication, we get

$m_1y=-x \ \text{and} \ m_2y=-x$

Add x on both sides of the equation, we get

$x+m_1y=0 \ \text{and} \ x+m_2y=0$

Therefore, the joint equation of the line is

$\left(x+m_{1} y\right)\left(x+m_{2} y\right)=0$

$x^2+m_2xy+m_1xy+m_1m_2y^2=0$

$x^{2}+\left(m_{1}+m_{2}\right) x y+m_{1} m_{2} y^{2}=0$

Substitute (1) and (2), we get

$$x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0$

To make the denominator same, multiply and divide first term by b.

$$ \frac{b}{b} x^{2}+\left(\frac{-2 h}{b}\right) x y+\left(\frac{a}{b}\right) y^{2}=0$

$$\frac{bx^2-2hxy+ay^2}{b} = 0$

Do cross multiplication, we get

$b x^{2}-2 h x y+a y^{2}=0$

Hence equation of the pair of lines perpendicular to the lines given equation is $b x^{2}-2 h x y+a y^{2}=0$ .

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