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Rudik [331]
3 years ago
6

Trigonometric identities assignment 4

Mathematics
1 answer:
kykrilka [37]3 years ago
6 0

Answer:

<em>Answer: choice D.</em>

Step-by-step explanation:

<u>Trigonometric Functions</u>

We need to recall some properties and definitions of the trigonometric functions.

\displaystyle \tan \theta=\frac{\sin\theta}{\cos\theta}

\displaystyle \sec \theta=\frac{1}{\cos\theta}

We need to simplify the following expression:

5\sin\theta\sec\theta

Substituting the identity for the secant:

\displaystyle 5\sin\theta\sec\theta=5\sin\theta.\frac{1}{\cos\theta}

Operating:

\displaystyle 5\sin\theta\sec\theta=5\frac{\sin\theta}{\cos\theta}

Substituting the definition of the tangent:

\displaystyle 5\sin\theta\sec\theta=5\tan\theta

Correct choice D.

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find the equation of a cubic function whose graph passes through points (3,0) and (1,4) and is tangent to x-axis at the origin
Tomtit [17]

Answer:

y = -2x^2(x - 3)

Step-by-step explanation:

<em><u>Preliminary Remark</u></em>

If a cubic is tangent to the x axis at 0,0

Then the equation must be related to y = a*x^2(x - h)

<em><u>(3,0)</u></em>

If the cubic goes through the point (3,0), then the equation will become

0 = a*3^2(3 - h)

0 = 9a (3 - h)

0 = 27a - 9ah

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<em><u>From the second point, we get</u></em>

4 = ax^2(x - 3)

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<em><u>Answer</u></em>

y = -2x^2(x - 3)

 

3 0
3 years ago
PLSSS HELP ME WITH MATH
labwork [276]

Step-by-step explanation:

always remember, equations and inequalities are like a scale with 2 pans (left and right).

a variable is like a placeholder, an empty spot on the pans. and the expressions on both sides are kind of the conditions for the things to put in these empty spots so that the status of the scale stays unchanged.

for an equation the scale has to stay balanced all the time.

and in an inequality one side is heavier that the other (it normally does not matter how much heavier) with "<" or ">" signs, or the scale can be also balanced with "<=" or ">=" signs. just the other side must never be heavier.

A

so, in our case here

2x - 5 = 3

yes, the solution (the ONLY solution, actually) is x = 4. no other value of x allows the scale to be balanced.

2x - 5 >= 3

well, since it is a ">=" sign, we can be lazy and treat it like the equation above, and x = 4 is therefore a valid solution for the inequality too.

but so is every other value of x that makes the left side "heavier". what about e.g. 5 ?

2×5 - 5 >= 3

10 - 5 >= 3

5 >= 3

true, great ! so, e.g. x = 5 is also a valid solution.

what else ?

let's simplify the inequality

2x - 5 >= 3

2x >= 8

x >= 4

so, really every value of x that is greater or equal to 4 is a valid solution for the inequality.

B

-2x - 5 = 3

yes, the solution (the ONLY solution, actually) is x = -4. no other value of x allows the scale to be balanced.

-2x - 5 >= 3

well, since it is a ">=" sign, we can be lazy and treat it like the equation above, and x = -4 is therefore a valid solution for the inequality too.

but so is every other value of x that makes the left side "heavier". what about e.g. -5 ?

-2×-5 - 5 >= 3

remember, -×- = +

10 - 5 >= 3

5 >= 3

true, great ! so, e.g. x = -5 is also a valid solution.

what else ?

let's simplify the inequality

-2x - 5 >= 3

-2x >= 8

x <= -4

a multiplication or division by a negative value flips the inequality sign, because such an operation makes a light weight heavy and a heavy weight light.

so, really every value of x that is less or equal to -4 is a valid solution for the inequality.

3 0
2 years ago
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