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bija089 [108]
3 years ago
11

What is the domain of the function y=3 in x graphed below?

Mathematics
2 answers:
Semenov [28]3 years ago
4 0

Answer:

domain is x>0

Step-by-step explanation:

y=3ln(x)

domain is the set of x values for which the function is defined

ln(negative value ) is undefined

So we take positive values for x because we have ln(x)

From the graph we can see that the given equation has a graph for x values greater than 0

the graph goes close to 0 but does not crosses y axis

So domain is x>0

Jet001 [13]3 years ago
3 0
What is the domain of the function y=3 in x graphed below?

Domain is x>0, ln (0) is undefined and there is no negative ln.
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After one hour, the hare had finished 2/3 of a 100-yard race. In that same time, the tortoise had finished 42 3/4 yards. How muc
andriy [413]

Answer:

23 11/12 yards or 23.9166 yards

Step-by-step explanation:

2/3 of 100 yard race is equivalent to 66 2/3 yards. Tge difference between the above and 42 3/4 yards covered by tortoise will be

66 2/3- 42 3/4= 23 11/12 yards or 23.91666 yards

5 0
3 years ago
4y-8-2y+5=0 help please
tia_tia [17]
4y - 8 - 2y + 5 = 0
2y - 3 = 0
2y = 3
y = 3/2
4 0
3 years ago
The region bounded by y=x^2+1, y=x, x=-1, x=2 with square cross sections perpendicular to the x-axis.
VLD [36.1K]

Answer:

The bounded area is 5 + 5/6 square units. (or 35/6 square units)

Step-by-step explanation:

Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)

Such that f(x) > g(x) in the given interval.

This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).

We want to find the area bounded by:

f(x) = y = x^2 + 1

g(x) = y = x

x = -1

x = 2

To find this area, we need to f(x) - g(x) between x = -1 and x = 2

This is:

\int\limits^2_{-1} {(f(x) - g(x))} \, dx

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx

We know that:

\int\limits^{}_{} {x} \, dx = \frac{x^2}{2}

\int\limits^{}_{} {1} \, dx = x

\int\limits^{}_{} {x^2} \, dx = \frac{x^3}{3}

Then our integral is:

\int\limits^2_{-1} {(x^2 + 1 - x)} \, dx = (\frac{2^3}{2}  + 2 - \frac{2^2}{2}) - (\frac{(-1)^3}{3}  + (-1) - \frac{(-1)^2}{2}  )

The right side is equal to:

(4 + 2 - 2) - ( -1/3 - 1 - 1/2) = 4 + 1/3 + 1 + 1/2 = 5 + 2/6 + 3/6 = 5 + 5/6

The bounded area is 5 + 5/6 square units.

3 0
2 years ago
Please help I think these are the right answers!!!
Papessa [141]

We performed the following operations:

f(x)=\sqrt[3]{x}\mapsto g(x)=2\sqrt[3]{x}=2f(x)

If you multiply the parent function by a constant, you get a vertical stretch if the constant is greater than 1, a vertical compression if the constant is between 0 and 1. In this case the constant is 2, so we have a vertical stretch.

g(x)=2\sqrt[3]{x}\mapsto h(x)=-2\sqrt[3]{x}=-g(x)

If you change the sign of a function, you reflect its graph across the x axis.

h(x)=-2\sqrt[3]{x}\mapsto m(x)=-2\sqrt[3]{x}-1=h(x)-1

If you add a constant to a function, you translate its graph vertically. If the constant is positive, you translate upwards, otherwise you translate downwards. In this case, the constant is -1, so you translate 1 unit down.

5 0
2 years ago
Willow is 63 inches tall. Her brother jayden is 6 feet 5 inches tall. How many inxhes twller is jayden?
Shalnov [3]
Willow = 63in

Jayder = 6ft 5in

1 foot equals 12 inches

6x12=72

So 72 inches plus 5 inches equals 77

So now you subtract Willows height from Jaydens height


77 - 63 = 14

Jayden is 14 inches taller 
4 0
2 years ago
Read 2 more answers
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