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2 years ago

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If baby potatoes costs 1.62 and if green beans costs 2.54 and vera bought 2.625 pounds of potatoes and 1.5 pounds of green beans

. and if she uses a 10 dollar bill to pay what will her change be?

1 answer:

Svetradugi [14.3K]2 years ago

5 0

Vera's change will be $1.94

Baby potatoes= 1.62

Green beans= 2.54

Amount spent on potatoes= 1.62×2.625

= 4.25

Amount spent on beans= 2.54× 1.5

= 3.81

The change can be calculated as follows

= 3.81 + 4.25

= 8.06

10-8.06

= 1.94

Hence the change is $1.94

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Answer:

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Step-by-step explanation:

Data provided in the question:

a = −1 + √3i and b = 2

to prove:

$\frac{1}{a+b}=\frac{1}{a} + \frac{1}{b}$

Considering the LHS

⇒ $\frac{1}{a+b}$

substituting the value of a and b, we get

⇒ $\frac{1}{−1 + \sqrt3i+2}$

or

⇒ $\frac{1}{1 + \sqrt3i}$

on multiplying and dividing by conjugate ( 1 - √3i )

we get

$\frac{1}{1 + \sqrt3i}\times\frac{1 - \sqrt3i}{1 - \sqrt3i}$

or

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or

$\frac{1 - \sqrt3i}{1 + 3}$ (as (√i)² = -1 )

or

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Now,

considering the RHS

$\frac{1}{a} + \frac{1}{b}$

substituting the value of a and b, we get

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or

⇒ $\frac{2\times1 + ( -1 + \sqrt3i)\times1}{(-1 + \sqrt3i)\times2}$

or

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or

⇒ $\frac{1 + \sqrt3i}{(-1 + \sqrt3i)\times2}$

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we get

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1 - \sqrt3i}{-1 - \sqrt3i}$

or

$\frac{1 + \sqrt3i}{(−1 + \sqrt3i)\times2}\times\frac{-1( 1 + \sqrt3i)}{-1 - \sqrt3i}$

or

$\frac{(1 + \sqrt3i}^2\times(-1){((-1)^2 - (\sqrt3i)^2)\times2}$

or

$\frac{(1^2 + (\sqrt3i)^2+2(1)(\sqrt3i)\times(-1)}{(1 + 3)\times2}$

or

$\frac{(1 - 3 + 2\sqrt3i)\times(-1)}{(4)\times2}$

or

$\frac{(-2 + 2\sqrt3i)\times(-1)}{(4)\times2}$

or

$\frac{-2( 1 - 2\sqrt3i)\times(-1)}{(4)\times2}$

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$\frac{( 1 - 2\sqrt3i)}{(4)}$

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3 years ago

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Got it. Hope this will help.

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3 years ago

Which point is the solution to the following system of equations?

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The point (3, 2) is the solution to given system of equations

<em><u>Solution:</u></em>

Given that system of equations are:

$x^2 + y^2 = 13$ ------ eqn 1

$2x - y = 4$ ------- eqn 2

From eqn 2,

y = 2x - 4

Substitute y = 2x - 4 in eqn 1

$x^2 + (2x - 4)^2 = 13\\\\x^2 + 4x^2 + 16 - 16x = 13\\\\5x^2 -16x + 3 = 0$

Let us solve the above equation by quadratic formula,

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

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The discriminant $b^2 - 4ac>0$ so, there are two real roots.

$\begin{aligned}&x=\frac{16 \pm \sqrt{196}}{10}=\frac{16 \pm 14}{10}\\\\&x=\frac{16+14}{10} \text { or } \frac{16-14}{10}\\\\&x=\frac{30}{10} \text { or } x=\frac{2}{10}\\\\&x=3 \text { or } x=0.2\end{aligned}$

Substitute for x = 0.2 and x = 3 in 2x - y = 4

<em><u>when x = 3</u></em>

2(3) - y = 4

6 - y = 4

y = 2

<em><u>when x = 0.2</u></em>

2(0.2) - y = 4

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y = 0.4 - 4

y = -3.6

Thus Option D is correct The point is (3, 2)

7 0

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