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nadya68 [22]
3 years ago
8

6x/5 - x = x/15 - 10/3

Mathematics
2 answers:
Andrei [34K]3 years ago
7 0

Answer:

1the answers is

email is 3rd

Vadim26 [7]3 years ago
4 0

Answer:

x=-25

6x/5 - x = x/15 - 10/3

x/5 = x/15 - 10/3

3x = x - 50

3x - x = -50

2x = -50

x = -50/2

X=-25

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Is N² + N always an even number? Explain
qaws [65]

Answer:

Yes

Step-by-step explanation:

Yes.

Let's first start with even numbers. (N - even number)

Any even number squared, is an even number. Then, we add that squared even number to another even number which would give us an even number.

Now, let's see odd numbers.

Any odd number squared would be an odd number. When you add "N", it would be adding an odd number to an odd number, which gives you an even number.

OR

We can start by factoring the expression:

N^2+N=N(N+1)

This is essentially multiplying two consecutive numbers, which in turn means that one number has to be even, and one has to be odd. An odd number multiplied by an even number will always be even.

7 0
3 years ago
An object launched straight up at a speed of 29.4 meters per second has a height, h, in meters of h , t seconds after the object
Nostrana [21]

Answer:

h = 44.06 meters (maximum height)

the time the object takes to complete this whole path is 6 seconds, this is why the time at which the object reaches its maximum height will between 0 and 6 seconds

Step-by-step explanation:

To solve this question, we need to first recognize that this is a constant acceleration problem, specifically, it can be thought of as a projectile motion problem.

Recall, the equations of motion:

1) v^2 - v_0^2 = 2a(s - s_0)\\2) s = v_0^2  + \frac{1}{2} at^2\\3) v = v_0 + at

What do we already know?

  • v_0 = 29.4 ms^-1
  • The launch is straight up
  • a = -9.81 ms^-2 this is the gravitational acceleration g
  • s_0 = 0 m, since our reference point is at s = 0, (the ground)

We can use use the Eq(1):

we know that when any object is launched up, at maximum height its velocity is going to be zero, v = 0 ms^-2

v^2 - v_0^2 = 2a(s - s_0)\\0^2 - (29.4)^2 = 2(-9.81)(s- 0)\\s = 44.06 m

this is the maximum height!

Why does t have to between zero and six?

We can answer this using a bit visualization, if you think about the second equation

s = v_0 t - \frac{1}{2}at^2\\ s = 29.4t - 4.905t^2

this is the equation of the whole trajectory that object makes.

and if you solve this by making s = 0, you will get the times at which the object was at the ground. the times will be 0s and 5.99s.

so the amount of time the object takes to go through this whole path is 6 seconds and this why the object will only reach its maximum height in between this time interval.

hope this helps :)

5 0
3 years ago
Given that QRSTU=QXWVU, find
Mrac [35]
Angle UQR = Angle UQX since they are corresponding angles. Note how the top polygon reflects over the line QU to get the bottom polygon.

So because of this, and because angle UQX = 45, this means the answer is 45 degrees
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Anuta_ua [19.1K]
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Answer:

the answer is 4/5. The decimal does not repeat

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