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astraxan [27]
3 years ago
11

Prove that root6+root2 is irrational

Mathematics
1 answer:
Verdich [7]3 years ago
3 0

Answer:

It is irrational because the

\sqrt{6 }  \: and \:  \sqrt{2}

is not a perfect square

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Which of the lines graphed in the diagram represents the equation 6x + 4y = 8
Blizzard [7]
In order to easily discern which graph is a proper representation of 6x + 4y = 8, you first need to convert the equation to y = mx+ b, also known as slope-intercept form. Here's how you can do this:
6x + 4y = 8
4y = -6x + 8
y = -1.5x + 2
The +2 tells you that your line will intercept the vertical y-axis at (0, 2). This narrows it down to graphs a and d. Then, because you have a NEGATIVE number in front of your x (it's -1.5), you can tell that your graph will be going down as it moves from left to right. This leaves you with graph d as your answer!
6 0
3 years ago
Read 2 more answers
Solve the system of equations by graphing. -x+y = -2 y = -1​
Aleks04 [339]

Answer:

the solution is (9, -1).

Step-by-step explanation:

graph the two equations on a graphing calculator (i use a website called desmos) and the point of intersection is the solution.

i hope this helps! have a great day <3

4 0
3 years ago
The expression (7-4i)-(3-12i) simplified is
Marizza181 [45]

Answer:

8i + 4

Step-by-step explanation:

Add/subtract like terms:

7 - 3

= 4

-4i - (-12i)

-4i + 12i

= 8i

Add these together:

8i + 4

So, the simplified expression is 8i + 4

4 0
3 years ago
How many students will chos pepperoni
yawa3891 [41]
The answer will be 50%
4 0
3 years ago
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if h=-16t ²+160t represents the height of a rocket in T seconds after it was fired when will the rocket hit the ground?
makkiz [27]

Answer:

The time after which the rocket hit the ground is 5 seconds

Step-by-step explanation:

Given as :

The distance of the cover by rocket at the height h = - 16 t² + 160 t

Let the time after which rocket hit the ground = T seconds

Now, When the rocket hits the ground, then at that time, the velocity of the rocket becomes zero.

I.e velocity = \frac{\partial h}{\partial t} = 0

Or, v = \frac{\partial h}{\partial t} = 0

Now, v = \frac{\partial h}{\partial t}

Or, v = \frac{\partial (-16t^{2}+160t)}{\partial t}

or, v = - 32 t + 160

Now, ∵ velocity of rocket after reaching the ground becomes zero

So, v = - 32 t + 160 = 0

Or, 32 t = 160

Or, t = \frac{160}{32}

∴ t = 5 sec

So, The time after which the rocket hit the ground = 5 sec

Hence The time after which the rocket hit the ground is 5 seconds answer

6 0
3 years ago
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