2 years ago

HELP PLS, THANK YOUU

Mathematics

2 answers:

Scorpion4ik [409]2 years ago

6 0

1.) true

2.) false

3.) false

4.) true

5.) true

6.) False

7.) true

8.) false

Andrews [41]2 years ago

4 0

Answer:

1. True

2. False

3. False

4. True

5. True

6. False

7. True

8. False

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How do you subtract when there are zero's at the end? For ex. 1000-325 how'd you borrow from 1000?

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PLEASE HELP QUICK!! WILL OFFER 100 POINTS TO THE FIRST ONE THAT ANSWERS

Reil [10]

Given

$x+1 = \sqrt{7x+15}$

We have to set the restraint

$x+1\geq 0 \iff x \geq -1$

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

$(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0$

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

$x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}$

So, we have to impose

$x-7\geq 0 \iff x \geq 7$

Squaring both sides, we have

$(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0$

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

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