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damaskus [11]
3 years ago
14

Please Help 11th Grade Math ASAP It is really hard so I need someone really smart.

Mathematics
2 answers:
Zolol [24]3 years ago
6 0

Answer:

i am in 7th grade

Step-by-step explanation:

Alja [10]3 years ago
3 0
It’s 50>15
if you add 20+59 it’s concludes to 50>15 because you then multiple
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Find the value of x.<br> 6x+3(-3x+4)=13
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x= -1/3 or -0.3

Step-by-step explanation:

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What is the area of a square that has a side length of 4x
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A square with a side length of 4 would have an area of 16 because 4x4 equals 16.
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PLEASE HELP
Law Incorporation [45]
None of these are correct, however of the ones you listed, A would be the most likely to be correct. y is greater than 3... therefore making y a number greater than zero.
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Which situation is best represented by the following equation?
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D

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7 0
3 years ago
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This is for algebra 2 can someone tell me if I’m right and can someone explain how do you get a= ?
algol13

Answers:

a = 2

h = 3

k = -3

You have the correct h and k values. Nice work so far.

==========================================================

Explanation:

(h,k) is the vertex. So (h,k) = (3,-3) meaning h = 3 and k = -3

To get the value of 'a', we need to plug in some point on the red graph. Each point is of the form (x,y). We can pick any point we want that isn't the vertex.

Let's pick (0,3) which is the y intercept. I'm picking this because 0 is an easy number to work with

------------

y = a|x-h| + k

y = a|x-3| + (-3) .... plug in the h,k values

y = a|x-3| - 3

3 = a|0-3| - 3 .... plug in (x,y) = (0,3)

3 = a|-3| - 3

3 = a*3 - 3

3 = 3a - 3

3a-3 = 3

3a = 3+3 ... adding 3 to both sides

3a = 6

a = 6/3 ... divide both sides by 3

a = 2

The value a = 2 indicates that the parent function y = |x| has been stretched vertically by a factor of 2. So it is twice as tall as before. Then it has been shifted to place the vertex at (3,-3) as shown in the graph.

------------

You may be wondering why you can't pick on the vertex for (x,y)

Let's see what happens if we use (x,y) = (3,-3)

y = a|x-3| - 3

-3 = a|3-3| - 3 ... plug in (x,y) = (3,-3)

-3 = a|0| - 3 ... uh oh, we get 0 here

-3 = a*0 - 3

-3 = 0 - 3

-3 = -3

We get a true statement, which is nice, but it doesn't tell us anything about what the value of 'a' is. That 'a' term goes away entirely. So I avoided using x = 3 to prevent x-3 from being 0.

4 0
3 years ago
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