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3 years ago

Are 5y+(-4) and 4+(-5y) equivalent expressions

Mathematics

1 answer:

alexira [117]3 years ago

5 0

Answer:

no because the first one is equal to 6 and the other one is equal to -6

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a guy walks into a store and steals a $100 bill from the register without the owners knowldge. He then buys $70 worth of goods u

Elanso [62]

He lost $30 bc
Guy stole $100
Guy gave back $70
Owner gives hue $30 !
If that makes any sense ‍♀️

8 0

3 years ago

Complete the square to rewrite y-x^2-6x+14 in vertex form. then state whether the vertex is a maximum or minimum and give its co

ycow [4]

Answer:

$y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2$

And solving we have:

$y= x^2 -6x +9 + 14 -9$

$y= (x-3)^2 +5$

And we can write the expression like this:

$y-5 = (x-3)^2$

The vertex for this case would be:

$V= (3,5)$

And the minimum for the function would be 3 and there is no maximum value for the function

Step-by-step explanation:

For this case we have the following equation given:

$y= x^2 -6x +14$

We can complete the square like this:

$y= x^2 -6x +(\frac{6}{2})^2 +14 -(\frac{6}{2})^2$

And solving we have:

$y= x^2 -6x +9 + 14 -9$

$y= (x-3)^2 +5$

And we can write the expression like this:

$y-5 = (x-3)^2$

The vertex for this case would be:

$V= (3,5)$

And the minimum for the function would be 3 and there is no maximum value for the function

4 0

3 years ago

Please help me answer one of these!

babunello [35]

This seems to be referring to a particular construction of the perpendicular bisector of a segment which is not shown. Typically we set our compass needle on one endpoint of the segment and compass pencil on the other and draw the circle, and then swap endpoints and draw the other circle, then the line through the intersections of the circles is the perpendicular bisector.

There aren't any parallel lines involved in the above described construction, so I'll skip the first one.

2. Why do the circles have to be congruent ...

The perpendicular bisector is the set of points equidistant from the two endpoints of the segment. Constructing two circles of the same radius, centered on each endpoint, guarantees that the places they meet will be the same distance from both endpoints. If the radii were different the meets wouldn't be equidistant from the endpoints so wouldn't be on the perpendicular bisector.

3. ... circles of different sizes ...

[We just answered that. Let's do it again.]

Let's say we have a circle centered on each endpoint with different radii. Any point where the two circles meet will then be a different distance from one endpoint of the segment than from the other. Since the perpendicular bisector is the points that are the same distance from each endpoint, the intersection of circles with different radii isn't on it.

4. ... construct the perpendicular bisector ... a different way?

Maybe what I first described is different; there are no parallel lines.

8 0

3 years ago

Can someone help???

OLga [1]