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Zigmanuir [339]

1 year ago

14

Please help me!!!!!!!!!!!!!!!!!!!!!!!1

1 answer:

Ilya [14]1 year ago

8 0

Answer:

Step-by-step explanation:

I learned that in like 4th or 5th grade

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Jade says that the absolute value of a number and the opposite of a number is the same thing. Is Jade correct? Use examples to h

avanturin [10]

Answer:

No, the absolute value of a number is different from the opposite.

Step-by-step explanation:

For example, the absolute value of 4 is 4.

l4l = 4

But the opposite of 4 is -4.

Absolute value is ALWAYS positive, but the opposite of a positive number is a negative number.

8 0

2 years ago

Pls help! I am really stuck.<br> Find the domain of this graph.

Rudik [331]

Answer:

8

Step-by-step explanation:

The domain is from the points (-4,0) through (4,0).

6 0

2 years ago

Read 2 more answers

Identify the zeros of the function f(x) = 4x2 − 8x − 1 using the Quadratic Formula. HELP ASAP!!

forsale [732]

$1+\dfrac{\sqrt{5}}{2},1-\dfrac{\sqrt{5}}{2}$

Step-by-step explanation:

The given equation is $4x^{2}-8x-1$

Let $a$ be the coefficient of $x^{2}$

Let $b$ be the coefficient of $x$

Let $c$ be the constant.

Then the roots α,β for the equation $ax^{2}+bx+c$ are $\dfrac{-b+\sqrt{b^{2}-4ac} }{2a},\dfrac{-b-\sqrt{b^{2}-4ac} }{2a}$

So,α= $\frac{-b+\sqrt{b^{2}-4ac} }{2a}=\frac{8+\sqrt{64+16} }{8}=\frac{8+4\sqrt{5}}{8}=1+\frac{\sqrt{5}}{2}$

β= $\frac{-b-\sqrt{b^{2}-4ac} }{2a}=\frac{8-\sqrt{64+16} }{8}=\frac{8-4\sqrt{5}}{8}=1-\frac{\sqrt{5}}{2}$ .

So the roots are $1+\frac{\sqrt{5}}{2},1+\frac{\sqrt{5}}{2}$

5 0

2 years ago

What is the inverse and restricted domain of the equation 8x^2-3

kari74 [83]

Answer: $\bold{y=\pm \dfrac{\sqrt{2(x+3)}}{4},\qquad x\neq -3}$

<u>Step-by-step explanation:</u>

y = 8x² - 3 (Restriction: none - x is All Real Numbers)

The inverse is when you swap the x's and y's and then solve for y

x = 8y² - 3 <em>swapped the x and y</em>

x + 3 = 8y² <em>added 3 to both sides</em>

$\dfrac{x+3}{8}=y^2$ <em>divided both sides by 8</em>

$\sqrt{\dfrac{x+3}{8}}=\sqrt{y^2}$ <em>square rooted both sides</em>

$\pm \sqrt{\dfrac{x+3}{8}}=y$ <em>simplified</em>

$\pm \sqrt{\dfrac{x+3}{8}\bigg(\dfrac{2}{2}\bigg)}=y$ rationalized the denominator

$\pm \sqrt{\dfrac{2(x+3)}{16}}=y$ <em>simplified</em>

$\pm \dfrac{\sqrt{2(x+3)}}{4}=y$ <em>simplified</em>

<u>Restriction:</u>

The radical <em>(inside the square root sign)</em> cannot be negative

→ 2(x + 3) ≥ 0

x + 3 ≥ 0 <em>divided both sides by 2</em>

x ≥ -3 <em>subtracted 3 from both sides</em>

4 0

3 years ago

HELP DISTRIBUTIVE PROPERTY QUADRATICS PLZZZZZZZ

Vlad1618 [11]

1: x² + 10x + 25
2: x² - x -12
3: -2x² + 10x
4: -2x² + 9x + 5
5: n² - 4
---------------------------
1: x² + 14x + 49
2: x² - 14x + 49
3: 4n² - 20n + 25
4: 4n² + 20n +25
5: you just have to switch the signs

4 0

3 years ago