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melomori [17]
3 years ago
14

Graph 12345 then try to use them numbers to make a straight line on the graph

Mathematics
2 answers:
nirvana33 [79]3 years ago
6 0

Answer:

that cant be done

Step-by-step explanation:


m_a_m_a [10]3 years ago
6 0

Answer:

you cant

Step-by-step explanation:

you need points ex: (1, 1)

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Use the substitution method to solve each linear system. 4x - 3y = -13,-2x + y = 4
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Step-by-step explanation:

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I am not sure how to slove
notsponge [240]

First .... the limit of the given function is NOT EQUAL to 5 (it diverges so it has no limit).  There is a typo.  The 14 should be positive.

\lim_{x \to\ 7} \bigg(\dfrac{x^2-9x+14}{x-7}\bigg)=5

The precise definition of a limit is:

\text{If for every } \epsilon>0\text{ there exists a }\delta >0\text{ such that}\\|f(x)-L|

Given:

f(x) = \dfrac{x^2-9x+14}{x-7}\\\\L=5\\\\a=7\\\\\\|f(x)-L|

⇒ \epsilon = \delta

When ε = 0.1,     δ = 0.1

When ε = 0.01,   δ = 0.01

5 0
3 years ago
The rate at which rain accumulates in a bucket is modeled by the function r given by r(t)=10t−t^2, where r(t) is measured in mil
mars1129 [50]

Answer:

36 milliliters of rain.

Step-by-step explanation:

The rate at which rain accumluated in a bucket is given by the function:

r(t)=10t-t^2

Where r(t) is measured in milliliters per minute.

We want to find the total accumulation of rain from <em>t</em> = 0 to <em>t</em> = 3.

We can use the Net Change Theorem. So, we will integrate function <em>r</em> from <em>t</em> = 0 to <em>t</em> = 3:

\displaystyle \int_0^3r(t)\, dt

Substitute:

=\displaystyle \int_0^3 10t-t^2\, dt

Integrate:

\displaystyle =5t^2-\frac{1}{3}t^3\Big|_0^3

Evaluate:

\displaystyle =(5(3)^2-\frac{1}{3}(3)^3)-(5(0)^2-\frac{1}{3}(0)^3)=36\text{ milliliters}

36 milliliters of rain accumulated in the bucket from time <em>t</em> = 0 to <em>t</em> = 3.

4 0
3 years ago
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