Write an equation in point slope form that passes through the given point and has the given slope.

Another path in the community is 2.4 miles long. It has benches at 1/3 And 2/3 of the distance from beginning to end of the path

antoniya [11.8K]

Answer:

the benches are 0.8 and 1.6 miles from the ends of the path

Step-by-step explanation:

The first bench is 1/3 of the path's length from the beginning of the path:

(1/3)×(2.4 mi) = (2.4/3) mi = 0.8 mi

The second bench is 2/3 of the path's length from the beginning of the path:

(2/3)×(2.4 mi) = (2.4×2/3) mi = 4.8/3 mi = 1.6 mi

The benches are 0.8 mi and 1.6 mi from the beginning of the path.

_____

Of course, 0.8 mi from the beginning is 2.4-0.8 = 1.6 mi from the end, and vice-versa.

5 0

3 years ago

Teresa made $192 for 12 hours of work at the same rate how much would she make for 18 hours of work

Nesterboy [21]

Step-by-step explanation:

$192=12hours

$576=36hours

$96=6hours

$288=18hours

7 0

3 years ago

A man spends 2/7 of his $1900 salary on loans. How much many remain with him?

Alex17521 [72]

Answer:

$1357.14

Step-by-step explanation:

Given that 2/7 of his Salary is on loans

2/7(1900) = 542.86

Remaining amount = $1900-$542.86 = $1357.14

4 0

3 years ago

I need help with finding the answer to a) and b). Thank you!

shtirl [24]

Answer:

$\displaystyle \sin\Big(\frac{x}{2}\Big) = \frac{7\sqrt{58} }{ 58 }$

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\frac{3 \sqrt{58}}{58}$

$\displaystyle \tan\Big(\frac{x}{2}\Big)=-\frac{7}{3}$

Step-by-step explanation:

We are given that:

$\displaystyle \sin(x)=-\frac{21}{29}$

Where x is in QIII.

First, recall that sine is the ratio of the opposite side to the hypotenuse. Therefore, the adjacent side is:

$a=\sqrt{29^2-21^2}=20$

So, with respect to x, the opposite side is 21, the adjacent side is 20, and the hypotenuse is 29.

Since x is in QIII, sine is negative, cosine is negative, and tangent is also negative.

And if x is in QIII, this means that:

$180$

So:

$\displaystyle 90 < \frac{x}{2} < 135$

Thus, x/2 will be in QII, where sine is positive, cosine is negative, and tangent is negative.

1)

Recall that:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\pm\sqrt{\frac{1 - \cos(x)}{2}}$

Since x/2 is in QII, this will be positive.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{1 + 20/29}{2}$

Simplify:

$\displaystyle \sin\Big(\frac{x}{2}\Big)=\sqrt{\frac{49/29}{2}}=\sqrt{\frac{49}{58}}=\frac{7}{\sqrt{58}}=\frac{7\sqrt{58}}{58}$

2)

Likewise:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =\pm \sqrt{ \frac{1+\cos(x)}{2} }$

Since x/2 is in QII, this will be negative.

Using the above information, cos(x) is -20/29. Therefore:

$\displaystyle \cos \Big( \frac{x}{2} \Big) =-\sqrt{ \frac{1- 20/29}{2} }$

Simplify:

$\displaystyle \cos\Big(\frac{x}{2}\Big)=-\sqrt{\frac{9/29}{2}}=-\sqrt{\frac{9}{58}}=-\frac{3}{\sqrt{58}}=-\frac{3\sqrt{58}}{58}$

3)

Finally:

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

Therefore:

$\displaystyle \tan\Big(\frac{x}{2}\Big)=\frac{7\sqrt{58}/58}{-3\sqrt{58}/58}=-\frac{7}{3}$

5 0

3 years ago

Help<br> plz (A) 4.00<br> (B) 0.25<br> (C) 1.00<br> (D)0.10

Mademuasel [1]

Answer:

B. 0.25

Step-by-step explanation:

The constant proportionality is 0.25 because the total cost is 0.25 or 1/4 of the number of tickets sold.

8 0

3 years ago

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