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Tresset [83]
1 year ago
12

Having trouble solving this problem from my ACT prep bookIt is trigonometry

Mathematics
1 answer:
Reptile [31]1 year ago
5 0

We need to convert 135.28º in degree, minute, and second measure.

In order to do so, notice that:

135.28\degree=135\degree+0.28\degree

Also,

1\degree=60^{\prime}

Thus, we have the proportion:

<em>degrees minutes</em>

<em> 1 60</em>

<em> 0.28 x</em>

Then:

x=0.28\cdot60=16.8

Thus:

0.28\degree=16.8^{\prime}=16^{\prime}+0.8^{\prime}

Now, notice that:

1^{\prime}=60^{\doubleprime}

Thus, we have the proportion:

<em>minutes seconds</em>

<em> 1 60</em>

<em> 0.8 y</em>

Then:

y=0.8\cdot60=48

Therefore:

135.28\degree=135\degree16^{\prime}48^{\doubleprime}^{}

And the first option is correct.

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Answer:

odd

Step-by-step explanation:

Just so you know there are shortcuts for determining if a polynomial function is even or odd. You just to make sure you use that x=x^1 and if you have a constant, write it as constant*x^0 (since x^0=1)

THEN!

If all of your exponents are odd then the function is odd

If all of your exponents are even then the function is even

Now you have -4x^3+4x^1

3 and 1 are odd it is an odd function

This a short cut not the legit algebra way

let me show you that now:

For it to be even you have f(-x)=f(x)

For it be odd you have f(-x)=-f(x)

If you don't have either of those cases you say it is neither

So let's check

plug in -x  -4(-x)^3+4(-x)=-4*-x^3+-4x=-4x^3+-4x

that's not the same so not even

with if we factor out -1 .... well if we do that we get -(4x^3+4x)=-f(x)

so it is odd.

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3 years ago
Cody pays $657 for six months of guitar lessons. He pays the same amount for lessons each month. Which of the following is the b
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a fish is 73 feet below sea level. It then swims 14 feet toward the surface. How many feet below sea level is the fish how?
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Two telephone calls come into a switchboard at random times in a fixed one-hour period. Assume that the calls are made independe
svp [43]

Answer:

0.25

Step-by-step explanation:

From the given information:

Since the times are uniform distributed over the one hour period (0,1):

Then;

f_1(y_1) = 1

f_2(y_2) = 1

So, Y_1 and Y_2 are independent; then:

f(y_1|y_2) = f_1(y_1)f_2(y_2)  \\ \\ = 1(1) = 1

P(Y_1\le 0.5 Y_2 \le 0.5) = \int ^{0.5}_{0} \int ^{0.5}_{0} f(y_1,y_2) dy_{2}dy_{1}

\implies  \int ^{0.5}_{0} \int ^{0.5}_{0} (1)  dy_{2}dy_{1}

= \dfrac{1}{4}

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2 years ago
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