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3 years ago

15

What variable do you eliminate?

1 answer:

puteri [66]3 years ago

7 0

It is A

Hope this helps :3

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What ideas do you need to remind yourself when working with taxes and discounts

azamat

Answer:

remember to read carefully

Step-by-step explanation:

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2 years ago

EMERGENCY (NEED HEP NOW!!!)

diamong [38]

Answer:

x=-2.25,y=-2.25

Step-by-step explanation:

The given system is

$4x - y = - 9 \\ 3x - 3y = 0$

We want to solve the system for x and y that makes the system true.

Let us multiply the first equation by 3 while maintaining the second equation:

$12x -3 y = - 27 \\ 3x - 3y = 0$

Subtract the current second equation from the first one.

$12x = - 27$

Divide through by 12:

$x = - \frac{27}{12} = - 2.25$

From the second equation we have:

$3x = 3y$

This implies that:

$x = y = - 2.25$

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3 years ago

For 30 minutes you do a combination of walking and jogging. At the end of your workout your pedometer displays a total of 2.5 mi

yKpoI14uk [10]

Walking for 10 minutes

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2 years ago

Find y<br>Solve the simultaneous <br>equations<br><br><br>3x + 2y = 27 3x - Y = 9<br>

lutik1710 [3]

Answer:

$y = 6$

Step-by-step explanation:

To find $y$ , we need to eliminate $x$ in this system of equations:

$3\cdot x + 2\cdot y = 27$ (1)

$3\cdot x - y = 9$ (2)

From (1) and (2):

$x = \frac{27-2\cdot y}{3}$

$x = \frac{9+y}{3}$

Then, we equalize both expressions and solve for $y$ :

$\frac{27-2\cdot y}{3} = \frac{9+y}{3}$

$27-2\cdot y = 9 + y$

$3\cdot y = 18$

$y = 6$

6 0

2 years ago

Which of the following have the property that a(x)=a−1(x)? I. y=x II. y=1/x III.y=x^2 IV. y=x^3 A. I and II, only B. IV, only C.

valentina_108 [34]

Answer:

<em>Correct answer:</em>

<em>A. I and II</em>

<em></em>

Step-by-step explanation:

First of all, let us have a look at the steps of finding inverse of a function.

1. Replace y with x and x with y.

2. Solve for y.

3. Replace y with $f^{-1}(x)$

Given that:

$I.\ y=x \\II.\ y=\dfrac{1}x \\III.\ y=x^2 \\IV.\ y=x^3$

Now, let us find inverse of each option one by one.

I. y = x, a(x) = x

Replacing y with and x with y:

x = y

x = $a^{-1}(x)$ = $a(x)$ Hence, I is true.

II. $y =\dfrac{1}{x}$

Replacing y with and x with y:

$x =\dfrac{1}{y}$

$x=\dfrac{1}{a^{-1}(x)}$

$\Rightarrow a^{-1}(x) = \dfrac{1}{x}$

$a^{-1}(x)$ = $a(x)$ Hence, II is true.

III. $y =x^{2}$

Replacing y with and x with y:

$x =y^{2}\\\Rightarrow y = \sqrt x\\\Rightarrow a^{-1}(x) = \sqrt{x} \ne a(x)$

Hence, III is not true.

IV. $y =x^{3}$

Replacing y with and x with y:

$x =y^{3}\\\Rightarrow y = \sqrt[3] x\\\Rightarrow a^{-1}(x) = \sqrt[3]{x} \ne a(x)$

Hence, IV is not true.

<em>Correct answer:</em>

<em>A. I and II</em>

<em></em>

4 0

2 years ago