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Tamiku [17]
3 years ago
13

Pls help due soon ty

Mathematics
2 answers:
Feliz [49]3 years ago
5 0
The first table : y=4x
the second table: y=1.5x
the third table: y= 1.25x

if you want help on how to do it: plug each of those equations into the “desmos” graph! from looking at those table points (x,y) you can find which equation matches the table

for instance on table one: I put y=4x into desmos then looked at graph one, I chose the point (2,8) I went to 2 on the x axis and moved my finger up 8 and that point was directly on the line meaning that table matches the equation. be sure to also do the same thing with the other points to double check!
My name is Ann [436]3 years ago
3 0

Answer:

The first table matches with the last equation.

The second table matches with the first equation.

The last table matches with the second equation.

Step-by-step explanation:

What I did was plug in the number in the x section for each table to see if the result of x times the number such as 4 resulted in the same number, y, that was paired with the x number based on the table.

Ex: Using table 1,

y=4x

y=4(2) --> I substituted the x for row 1 x in table 1

y=8

Check:

The first table shows that for row 1, the y was 8 so our match was correct.

Hope this helps!

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Solve the inequality 2x>30+5/4x
insens350 [35]

Answer:

Step-by-step explanation:

2x > 30+\frac{5}{4x} \\2x-\frac{5}{4x} > 30\\\frac{8x^2-5}{4x} > 30\\case~1\\if~x > 0\\8x^2-5 > 120x\\8x^2-120x > 5\\x^2-15x > \frac{5}{8} \\adding~(-\frac{15}{2} )^2~to~both~sides\\(x-\frac{15}{2} )^2 > \frac{5}{8}+\frac{225}{4} \\(x-\frac{15}{2} )^2 > \frac{455}{8} \\x-\frac{15}{2} < -\sqrt{\frac{455}{8} }  \\x < \frac{15}{2}-\sqrt{\frac{455}{8} } \\or~x < 0\\rejected~as~x > 0

x-\frac{15}{2} > \sqrt{\frac{455}{8} } \\x > \frac{15}{2} +\sqrt{\frac{455}{8} }

case~2

if~x < 0\\8x^2-5 < 120x\\8x^2-120x < 5\\x^2-15x < \frac{5}{8} \\adding~(-\frac{15}{2} )^2\\(x-\frac{15}{2} )^2 < \frac{5}{8} +(-\frac{15}{2} )^2\\|x-\frac{15}{2} | < \frac{5+450}{8} \\-\sqrt{\frac{455}{8} } < x-\frac{15}{2} < \sqrt{\frac{455}{8} } \\\frac{15}{2} -\sqrt{\frac{455}{8} } < x < \frac{15}{2} +\sqrt{\frac{455}{8} } \\but~x < 0\\7.5-\sqrt{\frac{455}{8} } < x < 0

8 0
1 year ago
Find the circumference of a circle with a radius of 8 inches round to the nearest 10th
Mkey [24]

Is the answer to your question 9?

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Is 44.11 greater than 44.5
Aleks04 [339]
No because I'll teach you a trick. Put a zero after the 5.   so now you have 44.11 and 44.50 . So 44.5 is bigger! 
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3 years ago
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Lady_Fox [76]
Just  matter of combining like terms
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- 4x - 7 <===
3 0
3 years ago
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