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ExtremeBDS [4]
3 years ago
8

Plllzz hellpppp meeeee thhhhxx

Mathematics
2 answers:
Citrus2011 [14]3 years ago
7 0
I also think it’s x=7
Naddik [55]3 years ago
6 0
Most likely it’s x=7 unless there are other options you didn’t show
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(3x-8)(2xsquared4x-9) multiply and write the result in descending order
RSB [31]
Hello there, you can proceed with the multiplication of these factors using the FOIL method (which stands for first, outer, inner and last) multiply I assume that by descending order, you mean by decreasing power. If it were so, the answer is 6(x^3)12(x^2)-16(x^2)32x-27x+72. 
8 0
3 years ago
120% of which is 480. I write what i know but i dont know what going on next, maybe I write wrong haha Improve me ​
Ilia_Sergeevich [38]
I’m pretty sure The answer is 1.2
8 0
2 years ago
Mary looks over reports on four of her workers. jack made 30 baskets in 5 hours. rudy made 32 baskets in 8 hours. sam made 40 ba
Elena L [17]
Well let's do some maths if each made a certain amount of baskets an hour you could divide them and figure out how many baskets were made per hour.
jack = 30/5 = 6 baskets per hour
Rudy = 32/8 = 4 baskets per hour
Sam = 40/12 =3.333 baskets per hour
Walter = 22/4 = 5.5 baskets per hour
according to this Jack had the greatest productivity
6 0
3 years ago
Mackenzie has one candy bar to share equally with her best friend. What portion of the candy bar does Mackenzie give to her best
ExtremeBDS [4]

Answer:1/2

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
The vertex of the parabola would be located at which coordinates<br> -6,3<br> 6,3<br> 3,-6<br> 3,6
Artemon [7]
<h3><u>Explanation</u></h3>
  • Given Equation

y = 4 {(x - 6)}^{2}  + 3

  • Vertex Form

y = a {(x - h)}^{2}  + k

where a-term determines how wide/narrow/upward or downward of the graph.

h-term determines the changes of graph for x-axis

k-term determines the changes of graph for y-axis.

The vertex of parabola is at (h,k).

y = 4 {(x - ( + 6)})^{2}  + ( + 3)

Therefore, h = 6 and k = 3.

<h3><u>Answer</u></h3>

<u>\large \boxed{(6,3)}</u>

7 0
2 years ago
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