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zalisa [80]
2 years ago
8

5m-3m+7n+2-5n please help​

Mathematics
2 answers:
ioda2 years ago
7 0

\\ \sf\bull\longmapsto 5m-3m+7n+2-5n

\\ \sf\bull\longmapsto 2m+7n-5n+2

\\ \sf\bull\longmapsto 2m+2n+2

\\ \sf\bull\longmapsto 2(m+n+1)

luda_lava [24]2 years ago
4 0

Answer:

\pmb{2(m+n+1)}

Step-by-step explanation:

  • 5m-3m+7n+2-5n

  • 2m+2n+2

  • 2(m+n+1)
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Numbers expressed using exponents are called .
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Answer:

numbers expressed using exponents are called powers

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2 years ago
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ABSOLUTE VALUE
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No the answer is B. t=6 or -6
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3 years ago
Which statements are true about the graph of the function f(x) = x2 – 8x + 5? Check all that apply.
statuscvo [17]

Answer:

A, D, E are true

Step-by-step explanation:

You have to complete the square to prove A.  Do this by first setting the function equal to 0, then moving the 5 to the other side.

x^2-8x=-5

Now we can complete the square.  Take half the linear term, square it, and add it to both sides.  Our linear term is 8 (from the -8x).  Half of 8 is 4, and 4 squared is 16.  So we add 16 to both sides.

(x^2-8x+16)=-5+16

We will do the addition on the right, no big deal.  On the left, however, what we have done in the process of completing the square is to create a perfect square binomial, which gives us the h coordinate of the vertex.  We will rewrite with that perfect square on the left and the addition done on the right,

(x-4)^2=11

Now we will move the 11 back over, which gives us the k coordinate of the vertex.

(x-4)^2-11=y

From this you can see that A is correct.

Also we can see that the vertex of this parabola is (4, -11), which is why B is NOT correct.

The axis of symmetry is also found in the h value.  This is, by definition, a positive x-squared parabola (opens upwards), so its axis of symmetry will be an "x = " equation.  In the case of this type of parabola, that "x = " will always be equal to the h value.  So the axis of symmetry is

x = 4, which is why C is NOT correct, either.

We can find the y-intercept of the function by going back to the standard form of the parabola (NOT the vertex form we found by completing the square) and sub in a 0 for x.  When we do that, and then solve for y, we find that when x = 0, y = 5.  So the y-intercept is (0, 5).

From this you can see that D is also correct.

To determine if the parabola has real solutions (meaning it will go through the x-axis twice), you can plug it into the quadratic formula to find these values of x.  I just plugged the formula into my graphing calculator and graphed it to see that it did, indeed, go through the x-axis twice.  Just so you know, the values of x where the function go through are (.6833752, 0) and (7.3166248, 0).  That's why you need the quadratic formula to find these values.

7 0
3 years ago
PLEASE HELP!!! I DONT HAVE ENOUGH POINTS TO POST MORE THAN ONCE!
Marat540 [252]

Answer:

-3.

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In the table, the y value goes down by -4 and then -2, for a total of -6.

We did this over a period of two units (To go from -1 to 1, we add 2).

-6/2 = -3.

-3 is the average rate of change over the interval.

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GIVING BRAINLIEST! A rectangular garden has a width of 10 feet and a length of 14 feet. A cement walkway is added around the out
Serjik [45]

Given:

Width of a garden = 10 feet

Length of the garden = 14 feet

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To find:

The width of the walkway.

Solution:

A cement walkway is added around the outside of the garden.

Let x be the width of the walkway.

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Area of a rectangle is

Area=length\times width

Area of garden and the walkway together is

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396=140+48x+4x^2

396=4(35+12x+x^2)

Divide both sides by 4.

99=35+12x+x^2

0=35-99+12x+x^2

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Splitting the middle term, we get

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Using zero product property, we get

(x+16)=0\text{ and }(x-4)=0

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Width of walkway cannot be negative. So, x=4.

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