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finlep [7]
3 years ago
8

Tim and Cynthia leave Cynthia's house at the same time. Tim drives north and Cynthia drives west. Tim's average speed is 10 mph

slower than Cynthia's. At the end of one hour, they are 70 miles apart. Find Cynthia's average speed. (Round your answer to the nearest tenth.)
Mathematics
1 answer:
RSB [31]3 years ago
6 0

Answer:

  54.2 mph

Step-by-step explanation:

Since the drivers leave at the same time and travel in directions 90° from each other, their separation speed can be found using the Pythagorean theorem. Let c represent Cynthia's average speed. Then (c-10) will represent Tim's average speed. Their combined separation speed is 70 miles per hour, so we can write ...

  70² = c² +(c -10)²

  4900 = 2c² -20c +100 . . . . eliminate parentheses

  c² -10c -2400 = 0 . . . . . . . . divide by 2; put in standard form

  (c -5)² -2425 = 0 . . . . . . . . . rearrange to vertex form

  c = 5 ±5√97 . . . . . . . . . . . . solve for c, simplify radical

  c ≈ 54.244 ≈ 54.2 . . . . . . . use the positive solution

Cynthia's average speed was 54.2 miles per hour.

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Suppose cos(x)= -1/3, where π/2 ≤ x ≤ π. What is the value of tan(2x). EDGE
AVprozaik [17]

Answer:

D

Step-by-step explanation:

We are given that:

\displaystyle \cos x = -\frac{1}{3}\text{ where } \pi /2 \leq x \leq \pi

And we want to find the value of tan(2<em>x</em>).

Note that since <em>x</em> is between π/2 and π, it is in QII.

In QII, cosine and tangent are negative and only sine is positive.

We can rewrite our expression as:

\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}

Using double angle identities:

\displaystyle  \tan(2x)=\frac{2\sin x\cos x}{\cos^2 x-\sin^2 x}

Since cosine relates the ratio of the adjacent side to the hypotenuse and we are given that cos(<em>x</em>) = -1/3, this means that our adjacent side is one and our hypotenuse is three (we can ignore the negative). Using this information, find the opposite side:

\displaystyle o=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}

So, our adjacent side is 1, our opposite side is 2√2, and our hypotenuse is 3.

From the above information, substitute in appropriate values. And since <em>x</em> is in QII, cosine and tangent will be negative while sine will be positive. Hence:

<h2>\displaystyle  \tan(2x)=\frac{2(2\sqrt{2}/3)(-1/3)}{(-1/3)^2-(2\sqrt{2}/3)^2}</h2>

Simplify:

\displaystyle  \tan(2x)=\frac{-4\sqrt{2}/9}{(1/9)-(8/9)}

Evaluate:

\displaystyle  \tan(2x)=\frac{-4\sqrt{2}/9}{-7/9} = \frac{4\sqrt{2}}{7}

The final answer is positive, so we can eliminate A and B.

We can simplify D to:

\displaystyle \frac{2\sqrt{8}}{7}=\frac{2(2\sqrt{2}}{7}=\frac{4\sqrt{2}}{7}

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7 0
3 years ago
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GenaCL600 [577]
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Work Shown:

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V = (1/3)*(12*10)*h

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V = 40h

Set this equal to the given pyramid volume 360 cubic inches and solve for h

40h = 360

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3 years ago
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Step-by-step explanation:

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8 0
3 years ago
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Keith_Richards [23]

Answer: An integer added to an integer is an integer, this statement is always true. A polynomial subtracted from a polynomial is a polynomial, this statement is always true. A polynomial divided by a polynomial is a polynomial, this statement is sometimes true. A polynomial multiplied by a polynomial is a polynomial, this statement is always true.

Explanation:

1)

The closure property of integer states that the addition, subtraction and multiplication is integers is always an integer.

If a\in Z\text{ and }b\in Z, then a+b\in Z.

Therefore, an integer added to an integer is an integer, this statement is always true.

2)

A polynomial is in the form of,

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Where a_n,a_{n-1},...,a_1,a_0 are constant coefficient.

When we subtract the two polynomial then the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

3)

If a polynomial divided by a polynomial  then it may or may not be a polynomial.

If the degree of numerator polynomial is higher than the degree of denominator polynomial then it may be a polynomial.

For example:

f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2

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f(x)=x^2-2x+5x-10 \text{ and } g(x)=x-2

Then \frac{g(x)}{f(x)}=\frac{1}{x^2+5}, which a not a polynomial.

Therefore, a polynomial divided by a polynomial is a polynomial, this statement is sometimes true.

4)

As we know a polynomial is in the form of,

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Where a_n,a_{n-1},...,a_1,a_0 are constant coefficient.

When we multiply the two polynomial, the degree of the resultand function is addition of degree of both polyminals and the resultant is also a polynomial form.

Therefore, a polynomial subtracted from a polynomial is a polynomial, this statement is always true.

3 0
3 years ago
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