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3 years ago

Find the slope of the line that contains the following points. R(-3,5), S(3,-2)

Mathematics

2 answers:

sesenic [268]3 years ago

6 0

Answer:

This can be solved using the slope formula.

(y2 - y1) / (x2 - x1) = m

m is the meaning of slope

Answer: m= -7/6

Step-by-step explanation:

Goryan [66]3 years ago

4 0

Answer:

-7/6

Step-by-step explanation:

var y/ var x

5 to -2 is variation y ----> -7

-3 to 3 is variation x ---> 6

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Hello,

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Find the value of the expression 6x + 3xy, when x = 2 and y = 4.

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<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

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3 years ago

An investment services company experienced dramatic growth in the last two decades. The following models for the company's reven

zlopas [31]

Answer: a) 138.32 and b) 35 years approx.

Step-by-step explanation:

Since we have given that

$R(t)=21.4e^{0.13t}\\\\C(t)=18.6e^{0.13t}$

So, Profit is given by

$Profit=R(t)-C(t)\\\\Profit=21.4e^{0.13t}-18.6e^{0.13t}\\\\Profit=e^{0.13t}(21.4-18.6)\\\\Profit=2.8e^{0.13t}$

Difference in years of 1990 and 2020=30

So, Profit becomes :

$P(30)=2.8e^{0.13\times 30}\\\\P(30)=2.8\times 49.40\\\\P(30)=138.32$

(b) How long before the profit found in part (a) is predicted to double? (Round your answer to the nearest whole number.) years after 1990.

So, profit doubles , we get :

$138.32\times 2=2.8e^{0.13t}\\\\276.65=2.8e^{0.13t}\\\\\dfrac{276.65}{2.8}=e^{0.13t}\\\\98.80=e^[0.13t}\\\\\ln 98.80=0.13t\\\\4.593=0.13t\\\\\dfrac{4..593}{0.13}=t\\\\35.33=t$

Hence, a) 138.32 and b) 35 years approx.

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3 years ago