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bagirrra123 [75]

3 years ago

7

The average gas mileage m in miles per gallon for a compact car is modeled by

1 answer:

Sauron [17]3 years ago

3 0

Answer:

In these functions translation is used.

The transformation means that vertex of function m(s) is translated 18 units downwards.

Step-by-step explanation:

We have been given that average gas mileage m in miles per gallon for a compact car is modeled by m(s) = - 0.015(s - 47)2 +33 and mileage for an SUV is modeled by my(s) = -0.015(s - 47)2 + 15.

We can see that both functions are parabolic and opens downwards. The vertex of function m(s) is (47,33), while vertex of function my(s) is (47,15). We can describe this change by translation.

This transformation means that function m(s) is translated 18 units downwards and average mileage of SUV is 18 miles/hour lesser than compact car at equal speeds.

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A.) which of these expressions is equivalent to 4 x x

Law Incorporation [45]

Answer:

a) The correct option is C: 4x

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Step-by-step explanation:

a) Usually when we have a real number multiplying a variable, we do not need to write the multiplication symbol.

So instead of writing:

4×x

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Then in this case the correct option is C.

b) We know that if we have the multiplication of A by n, this will be equivalent to add A n times.

Then if we have the sum of A, for example, 4 times, we have:

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And this can be written as 4*A

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Serjik [45]

Answer:

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Step-by-step explanation:

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Find \tan\left(\frac{17\pi}{12}\right)tan( 12 17π )tangent, left parenthesis, start fraction, 17, pi, divided by, 12, end frac

uranmaximum [27]

One way to do this is to notice

$\dfrac{17\pi}{12}=\dfrac\pi6+\dfrac{5\pi}4$

Then

$\tan\dfrac{17\pi}{12}=\tan\left(\dfrac\pi6+\dfrac{5\pi}4\right)=\dfrac{\tan\frac\pi6+\tan\frac{5\pi}4}{1-\tan\frac\pi6\tan\frac{5\pi}4}$

We have

$\tan\dfrac\pi6=\dfrac{\sin\frac\pi6}{\cos\frac\pi6}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}$

and since $\tan x$ has a period of $\pi$ ,

$\tan\dfrac{5\pi}4=\tan\left(\pi+\dfrac\pi4\right)=\tan\dfrac\pi4=1$

and so

$\tan\dfrac{17\pi}{12}=\dfrac{\frac1{\sqrt3}+1}{1-\frac1{\sqrt3}}=\dfrac{1+\sqrt3}{\sqrt3-1}=2+\sqrt3$

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3 years ago