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AURORKA [14]
2 years ago
10

You have 1.5

Mathematics
1 answer:
Alex787 [66]2 years ago
3 0
3

Please mark as brainliest
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PLS Answer these questions, I have a test!
Naya [18.7K]

Answer:

D, 120 , In order obtuse acute right,

Step-by-step explanation:

the fourth one dosent have options so I cant answer that ( It can be anything other than ZMD and DMZ)

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3 years ago
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Which of the following rational functions is graphed below?
podryga [215]

Answer:

c

Step-by-step explanation:

Oblique Asymptotes

R(x) will have oblique asymptote if it can be represented in the form

1 / (Q)x

4 0
1 year ago
Four pairs of shoes cost $80 what is the cost of seven pairs of shoes
anzhelika [568]

80 divided by 4 is 20

$20 times 7 shoes is

$140

Hope this helps!

5 0
3 years ago
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A particle is moving along the curve y= 4sqrt(5x+11) . As the particle passes through the point (5,24) , its -coordinate increas
kupik [55]

The rate of change of the distance from the particle to the origin at this instant is 3 units per second.

<h3>What is the rate of change?</h3>

The instantaneous rate of change is the rate of change at a particular instant.

A particle is moving along the curve

y= 4\sqrt{5x+11} .

The rate of change of y is given as:

dy / dx = 2

by differentiating both sides,

\dfrac{dy}{dx} = 4\dfrac{1}{2\sqrt{5x+11} }  5\dfrac{dx}{dt} \\\\\dfrac{dy}{dx} = \dfrac{10}{\sqrt{5x+11} }\dfrac{dx}{dt}\\\\

From the question, we have:

(x, y) =  (5,24)

Substitute 5 for x and dy / dx = 2

\dfrac{dy}{dx} = \dfrac{10}{\sqrt{5x+11} }\dfrac{dx}{dt}\\\\\\5= \dfrac{10}{\sqrt{5(5)+11} }\dfrac{dx}{dt}\\\\\\\sqrt{5(5)+11} = 2\dfrac{dx}{dt}\\\\\\\dfrac{dx}{dt} = 6/ 2 = 3

Hence, the rate of change of the distance from the particle to the origin at this instant is 3 units per second.

Read more about rates of change at:

brainly.com/question/13103052

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8 0
2 years ago
FIVE STARS AND BRAINLIEST TO CORRECT ANSWER
lianna [129]
Simply take the derivative of s(t),
s'(t) = -6

Since the velocity is constant, v(t) = s'(t) = -6 at any given time t.
6 0
3 years ago
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