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horrorfan [7]
2 years ago
8

Help help please please

Mathematics
1 answer:
gtnhenbr [62]2 years ago
8 0

✽ Hello there! ✽

4x+9<21  

Move all the constants (numbers) to the right, using the opposite                         operation:

4x<21-9

4x<12

Divide both sides by 4 to isolate x:

x<3

<h2>Therefore, x<3 ✔︎</h2>

Hope this helps!

~Just a felicitous girlie

#HaveASplendidDay

SilentNature

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How many and what type of solutions does the equation have?
Natali [406]
Let's solve the equation 2k^2 = 9 + 3k
First, subtract each side by (9+3k) to get 0 on the right side of the equation
2k^2 = 9 + 3k
2k^2 - (9+3k) = 9+3k - (9+3k)
2k^2 - 9 - 3k = 9 + 3k - 9 - 3k
2k^2 - 3k - 9 = 0

As you see, we got a quadratic equation of general form ax^2 + bx + c, in which a = 2, b= -3, and c = -9.
Δ = b^2 - 4ac
Δ = (-3)^2 - 4 (2)(-9)
Δ<u /> = 9 + 72
Δ<u /> = 81

Δ<u />>0 so the equation got 2 real solutions:
k = (-b + √Δ)/2a = (-(-3) + √<u />81) / 2*2 = (3+9)/4 = 12/4 = 3
AND
k = (-b -√Δ)/2a = (-(-3) - √<u />81)/2*2 = (3-9)/4 = -6/4 = -3/2

So the solutions to 2k^2 = 9+3k are k=3 and k=-3/2

A rational number is either an integer number, or a decimal number that got a definitive number of digits after the decimal point.

3 is an integer number, so it's rational.
-3/2 = -1.5, and -1.5 got a definitive number of digit after the decimal point, so it's rational.

So 2k^2 = 9 + 3k have two rational solutions (Option B).

Hope this Helps! :)
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3 years ago
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