I need help ASAP!!!<br><br><br> g(x) = x2

Okay, so, to find out if an equation has one solution, an infinite number of solutions, or no solutions, we must first solve the equation:

(a) 6x + 4x - 6 = 24 + 9x

First, combine the like-terms on both sides of the equal sign:

10x - 6 = 24 + 9x

Now, we need to get the numbers with the variable 'x,' on the same side, by subtracting, in this case:

10x - 6 = 24 + 9x
-9x. -9x
______________
X - 6 = 24

Now, we do the opposite of subtraction, and add 6 to both sides:

X - 6 = 24
+6 +6
_________
X = 30

So, this particular equation has one solution.

(a). One solution
_____________________________________________________

(b) 25 - 4x = 15 - 3x + 10 - x

Okay, so again, we combine the like-terms, on the same side of the equal sign:

25 - 4x = 25 - 2x

Now, we get the 2 numbers with the variable 'x,' to the same side of the equal sign:

25 - 4x = 25 - 2x
+ 2x + 2x
________________
25 - 2x = 25

Next, we do the opposite of addition, and, subtract 25 on each side:

25 - 2x = 25
-25 -25
___________
-2x = 0

Finally, because we can't divide 0 by -2, this tells us that this has an infinite number of solutions.

(b) An infinite number of solutions.

__________________________________________________

(c) 4x + 8 = 2x + 7 + 2x - 20

Again, we combine the like-terms, on the same side as the equal sign:

4x + 8 = 4x - 13

Now, we get the 'x' variables on the same side, again, and, we do that by doing the opposite of addition, which, is subtraction:

4x + 8 = 4x - 13
-4x -4x
______________
8 = -13

Finally, because there is no longer an 'x' or variable, we know that this equation has no solution.

(c) No Solution
_________________________________

I hope this helps!

4 0

3 years ago

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

kondaur [170]

$\text{Proof by induction:}$
$\text{Test that the statement holds or n = 1}$

$LHS = (3 - 2)^{2} = 1$
$RHS = \frac{6 - 4}{2} = \frac{2}{2} = 1 = LHS$
$\text{Thus, the statement holds for the base case.}$

$\text{Assume the statement holds for some arbitrary term, n= k}$
$1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2} = \frac{k(6k^{2} - 3k - 1)}{2}$

$\text{Prove it is true for n = k + 1}$
$RTP: 1^{2} + 4^{2} + 7^{2} + ... + [3(k + 1) - 2]^{2} = \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2} = \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$

$LHS = \underbrace{1^{2} + 4^{2} + 7^{2} + ... + (3k - 2)^{2}}_{\frac{k(6k^{2} - 3k - 1)}{2}} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1)}{2} + [3(k + 1) - 2]^{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2[3(k + 1) - 2]^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 2(3k + 1)^{2}}{2}$
$= \frac{k(6k^{2} - 3k - 1) + 18k^{2} + 12k + 2}{2}$
$= \frac{k(6k^{2} - 3k - 1 + 18k + 12) + 2}{2}$
$= \frac{k(6k^{2} + 15k + 11) + 2}{}$
$= \frac{(k + 1)[6k^{2} + 9k + 2]}{2}$
$= \frac{(k + 1)[6(k + 1)^{2} - 3(k + 1) - 1]}{2}$
$= RHS$

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.

3 0

3 years ago

Jimmy buys 5 notebooks that each cost the same amount and 1 binder that costs $7.50. The total cost is $18.75. How much does eac

Varvara68 [4.7K]

Answer:

A. $2.25

Step-by-step explanation:

If you subtract $7.50 from $18.75 you get $11.25, then divide that by 5 and you'll get $2.25.

Hope this was useful.

3 0

3 years ago

Read 2 more answers

I need help ASAP!!! g(x) = x2 - 2x2 + x g(3) =