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juin [17]
3 years ago
7

What is the slope between the points (5,-7) and (-7,1)?

Mathematics
1 answer:
SCORPION-xisa [38]3 years ago
6 0
I hope this helps you



(5,-7) x'=5 y'= -7


(-7,1) x"= -7 y"=1


slope. (x'-x")=y'-y"


slope. (5-(-7))= -7-1


slope. 12= -8


slope =-8/12


slope = -2/3
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Which contract costs more per month? An 18 month cable contract for $2070 or a 12 month contract at $1776? Which is the better d
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Contract A is cheaper because $115 per month is less than $148 per month

Step-by-step explanation:

To find the unit price we divide the total cost given by the total months given for each contract to find out the cost per month.

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$1776/12 months = $148 per month

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Answer:

The Answer is D. If the month is on the x axis then its gonna be above the horazontal line (x-axis) and if its a positive temerature then its gonna be in the quadrent 1.

Step-by-step explanation:

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<h3>How to determine the local minima?</h3>

The function is given as:

f\left(x\right)=\ \frac{\left(2x+3\right)^2\left(x\ -2\right)^5}{x^3\left(x-5\right)^2}

See attachment for the graph of the function f(x)

From the attached graph, we have the following minima:

Minimum = (-1.5, 0)

Minimum = (7.980, 609.174)

The above means that, the local minima are

(x, f(x)) = (-1.5, 0) and (7.980, 609.174)

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brainly.com/question/20394217

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7 0
1 year ago
Monique's son just turned 2 years old and is 34 inches tall. Monique heard that the average boy will grow approximately 2 5/8 in
tatuchka [14]

Answer:

The equation representing how old Monique son is \mathbf{a = 2 + \dfrac{8}{21}(q-34)}

Step-by-step explanation:

From the given information:

A linear function can be used to represent the constant growth rate of Monique Son.

i.e.

q(t) = \hat q \times t + q_o

where;

q_o = initial height of Monique's son

\hat q = growth rate (in)

t = time

So, the average boy grows approximately 2 5/8 inches in a year.

i.e.

\hat q = 2 \dfrac{5}{8} \ in/yr

\hat q =  \dfrac{21}{8} \ in/yr

Then; from the equation q(t) = \hat q \times t + q_o

34 = \dfrac{21}{8} \times 0 + q_o

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q(t) = \dfrac{21}{8} \times t + 34

Then:

Making t the subject;

q - 34 = \dfrac{21}{8} \times t

t = \dfrac{8}{21}(q-34)

and the age of the son  i.e. ( a (in years)) is:

a = 2 + t

So;

\mathbf{a = 2 + \dfrac{8}{21}(q-34)}

SO;

if q (growth rate) = 50 inches tall

Then;

\mathbf{a = 2 + \dfrac{8}{21}(50-34)}

\mathbf{a = 2 + \dfrac{8}{21}(16)}

a = 2 + 6.095

a = 8.095 years

a ≅ 8 years

i.e.

Monique son will be 8 years at the time Monique is 50 inches tall.

8 0
2 years ago
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