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Kazeer [188]
3 years ago
11

Graphing Natural Exponential Functions In Exercise,sketch the graph of the function.See Example 1.

Mathematics
1 answer:
nata0808 [166]3 years ago
6 0

Answer:

The graphic is attached

Step-by-step explanation:

Hi

x H(x)

0 6

≥5 lim┬(x→∞)⁡〖h(x)=5〗

≤5 lim┬(n→-∞)⁡〖h(x)= ∞〗

when x tends to infinity the function approaches five above (horizontal asymptote).

when x tends to infinity the function tends to infinity.

Success with your homework.

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In 1995, the moose population in a park was measured to be 1800. By 1998, the population was measured again to be 2400. If the p
Shalnov [3]

The linear equation for the variation of population is p = 200t + 1800.

<h3>What is an equation?</h3>

An equation is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

The formula for an equation in intercept form will be given as below:-

y = mx + c.

It is given that In 1995, the moose population in a park was measured to be 1800. By 1998, the population was measured again to be 2400.

Write the linear equation for population variation.

p = mt + c

m = Rate of change = ( 2400 - 1800 ) / 3 = 200, put the value in the equation.

p = 200t + c

Put the values in the equation to get the value of the constant.

2400 = 200 x 3 + c

c = 2400 - 600

c =1800

The equation for the linear variation of the population is,

p = 200t + 1800

Therefore, the linear equation for the variation of the population is p = 200t + 1800.

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2 years ago
Which expression is equivalent to the expression shown below?
OlgaM077 [116]

Answer:

5^{15}

Step-by-step explanation:

(5^3)^{9} \div 5^{12}=5^{3 \times 9} \div 5^{12}=5^{27} \div 5^{12}=5^{27-12}=5^{15}

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The answer would be B i think .

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