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Maurinko [17]
3 years ago
7

Write an expression with an exponent that has a value between 0 and 1

Mathematics
2 answers:
leva [86]3 years ago
8 0

Answer:

(1/2)^2, (2/5)^3

Step-by-step explanation:

First, let's think in a couple of examples

(\frac{1}{2})^2 = \frac{1^2}{2^2} = \frac{1}{4}

which is between 0 and 1

Another case is

(\frac{2}{5})^3 = \frac{2^3}{5^3} = \frac{8}{125}

which is between 0 and 1

In general any expression with this form

(\frac{n_1}{n_2})^{n_3}  

where n_1 < n_2, n_3 > 1 and n_1, n_2 and  n_3 are natural numbers; gives a result between 0 and 1.

Leokris [45]3 years ago
5 0

x^{1/2}

The above expression has an exponent between 0 and 1. This number is 1/2 or 0.5.

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Solve the following equation algebraically. Represent your answer in simplest radical form, if necessary.
maria [59]
x^2 + x = 30\\\\x^2 + x - 30=0\\\\x^2+6x-5x-30=0\\\\x(x+6)-5(x+6)=0\\\\(x+6)(x-5)=0\\\\x+6=0\ \ \ \ \ \ or\ \ \ \  \ \ x-5=0\\\\x=-6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=5
7 0
3 years ago
Differentiate f and find the domain of f. (enter the domain in interval notation.)f(x) = ln(x2 − 12x)derivative f '(x) = 2x−12x2
guajiro [1.7K]
F(x) = ln(x² - 12x)
The derivative is
f'(x)= \frac{2x-12}{x^{2}-12x}

f(x) is undefined when x² - 12x = x(x - 12) = 0
That is, when x = 0 or x = 12.
Therefore the domain is (-∞, 0)∪(0,12)∪(12, ∞)

Answer:
The derivative is
f'(x) =  \frac{2x-12}{x^{2}-12x}
The domain is
(-∞, 0)∪(0, 12)∪(12,∞)



3 0
3 years ago
A tunnel is built in form of a parabola. The width at the base of tunnel is 7 m. On
uysha [10]

Given:

The width at the base of parabolic tunnel is 7 m.

The ceiling 3 m from each end of the base there are light fixtures.

The height to light fixtures is 4 m.

To find:

Whether it is possible a trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel.

Solution:

The width at the base of tunnel is 7 m.

Let the graph of the parabola intersect the x-axis at x=0 and x=7. It means x and (x-7) are the factors of the height function.

The function of height is:

h(x)=ax(x-7)             ...(i)

Where, a is a constant.

The ceiling 3 m from each end of the base there are light fixtures and the height to light fixtures is 4 m. It means the graph of height function passes through the point (3,4).

Putting x=3 and h(x)=4 in (i), we get

4=a(3)((3)-7)

4=a(3)(-4)

\dfrac{4}{(3)(-4)}=a

-\dfrac{1}{3}=a

Putting a=-\dfrac{1}{3}, we get

h(x)=-\dfrac{1}{3}x(x-7)              ...(ii)

The center of the parabola is the midpoint of 0 and 7, i.e., 3.

The width of the truck is 4 m. If is passes through the center then the truck must m 2 m on the left side of the center and 2 m on the right side of the center.

2 m on the left side of the center is x=1.5.

A trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is possible if h(1.5) is greater than 2.8.

Putting x=1.5 in (ii), we get

h(1.5)=-\dfrac{1}{3}(1.5)(1.5-7)

h(1.5)=-(0.5)(-5.5)

h(1.5)=2.75

Since h(1.5)<2.8, therefore the trailer truck carrying cars is 4 m wide and 2.8 m high is going to drive through the tunnel is not possible.

6 0
3 years ago
A scaled factor of 2 is used to increase the triangle. What is the new area?
lesya [120]

Answer:b

Step-by-step explanation:b

8 0
3 years ago
Jake says adding 0 to an addendum does not change a sum. Is he correct
siniylev [52]
Nothing changes if you don't add anything. 

Example:

10+10=20
20+0=20
 
nothing changes.
5 0
3 years ago
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