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3 years ago

If u get a negative answer for exponents is it correct?

Mathematics

1 answer:

devlian [24]3 years ago

8 0

9514 1404 393

Answer:

maybe

Step-by-step explanation:

It can be. It depends on the problem.

The rules of exponents are ...

(a^b)/(a^c) = a^(b-c)

a^-b = 1/a^b

Clearly, for division problems if the denominator exponent is larger, the difference 'b-c' will be negative.

You recall that an exponent is the way we show repeated multiplication.

x·x·x = x³

In division, like factors cancel, so ...

$\dfrac{x\cdot x\cdot x}{x\cdot x}=\dfrac{x^3}{x^2}=\dfrac{x}{1}\cdot\dfrac{x\cdot x}{x\cdot x}=x^{3-2}=x^1=x$

Now, consider the same problem "upside down."

$\dfrac{x\cdot x}{x\cdot x\cdot x}=\dfrac{x^2}{x^3}=\dfrac{1}{x}\cdot\dfrac{x\cdot x}{x\cdot x}=x^{2-3}=x^{-1}=\dfrac{1}{x}$

You might be interested in

3.3^(2x+1)-103^x+1=0 need value of x

photoshop1234 [79]

Answer:

The value of $x$ is approximately -1.531.

Step-by-step explanation:

Let $3.3^{2\cdot x + 1}-103^{x+1} = 0$ , we proceed to solve this expression by algebraic means:

1) $3.3^{2\cdot x + 1}-103^{x+1} = 0$ Given

2) $3.3^{2\cdot x}\cdot 3.3 -103^{x}\cdot 103 = 0$ $a^{b}\cdot a^{c} = a^{b+c}$

3) $(3.3^{x})^{2}\cdot 3.3 -\left[\left( \sqrt{103} \right)^{2}\right]^{x}\cdot 103 = 0$ $(a^{b})^{c} = a^{b\cdot c}$

4) $(3.3^{x})^{2}\cdot 3.3 - \left[\left(\sqrt{103}\right)^{x}\right]^{2}\cdot 103 = 0$ $(a^{b})^{c} = a^{b\cdot c}$ /Commutative property

5) $\left[\left(\frac{3.3}{\sqrt{103}}\right)^{x}\right] ^{2}-\frac{103}{3.3} = 0$ Existence of multiplicative inverse/Definition of division/Modulative property/ $a^{b}\cdot a^{c} = a^{b+c}$

6) $\left(\frac{3.3}{\sqrt{103}} \right)^{2\cdot x}=\frac{103}{3.3}$ Existence of additive inverse/Modulative property/ $(a^{b})^{c} = a^{b\cdot c}$

7) $\log \left(\frac{3.3}{\sqrt{103}} \right)^{2\cdot x}=\log \frac{103}{3.3}$ Definition of logarithm.

8) $2\cdot x\cdot \log \left(\frac{3.3}{\sqrt{103}} \right)= \log \frac{103}{3.3}$ $\log_{b} a^{c} = c\cdot \log_{b} a$

9) $2\cdot x \cdot [\log 3.3-\log \sqrt{103}] = \log 103 - \log 3.3$ $\log_{b} \frac{a}{d}$

10) $x\cdot (2\cdot \log 3.3-\log 103) = \log 103 - \log 3.3$ $\log_{b} a^{c} = c\cdot \log_{b} a$ /Associative property

11) $x = \frac{\log 103-\log 3.3}{2\cdot \log 3.3-\log 103}$ Existence of multiplicative inverse/Definition of division/Modulative property

12) $x \approx -1.531$ Result

The value of $x$ is approximately -1.531.

4 0

3 years ago

A scientist dropped an object from a height of 200 feet. She recorded the height of the object in 0.5-second intervals. Her data

Reil [10]

The quadratic model is given by:
y = ax2 + bx + c
We evaluate three points to find the values of a, b and c.
We have then:
For (0, 200):
200 = a (0) 2 + b (0) + c
c = 200
For (0.5, 195):
195 = a * (0.5) ^ 2 + b * (0.5) +200
195-200 = a * (0.5) ^ 2 + b * (0.5)
-5 = 0.25a + 0.5b
For (1, 185):
185 = a * (1) ^ 2 + b * (1) +200
185-200 = a * (1) ^ 2 + b * (1)
-15 = a + b
Solving the system:
-5 = 0.25a + 0.5b
-15 = a + b
We have the following results:
a = -10
b = -5
Substituting:
y = -10x ^ 2 - 5x + 200
For x = 3 we have:
y = -10 (3) ^ 2 - 5 (3) + 200
y = - 90 - 15 + 200
y = 95
Answer:
the height at 3 seconds is:
y = 95 feet

3 0

3 years ago

Which expression is equivalent to

juin [17]

Answer:

Step-by-step explanation:

-2x - 8 - 3x - 8

= - 5x - 16

3 0

3 years ago

(29 PTS) Explain how the Quotient of Powers Property was used to simplify this expression. 3^4/9 = 3^2

Virty [35]

Step-by-step explanation:

d is the correct ans

see,

As the bases are equal we can cut the bas,

6 0

4 years ago

Read 2 more answers

ITS EASY BUT I DIDNT STUDY - PLZ HELP

SVEN [57.7K]

Answer:

The answer is b im doing a type limit.

6 0

3 years ago