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Mrac [35]
3 years ago
14

In the town of sleepy Oak. the fine for speeding ticket is $32.65 + s dollars. Where s is the miles per hour over the speed limi

t.
A. What is the fine for going 38.4 miles per hour in a 25-miles-per-hour school zone?

B. Mr. Taylor was fined $50.15 for speeding in the same school zone. How fast was he driving?
Mathematics
1 answer:
Mkey [24]3 years ago
5 0
A:  38.4 - 25 = 13.4
      13.4 + 32.65 = $46.05

B:  50.15 - 32.65 =  17.5
         17.5 + 25 =  42.5 mph
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Find the equation of the line passing
Ostrovityanka [42]

Answer:

  y = 4x -14

Step-by-step explanation:

The required answer form is slope-intercept form, so we need the slope and the y-intercept of the line.

The slope equation is ...

  m = (y2 -y1)/(x2 -x1)

  m = (10 -2)/(6 -4) = 8/2 = 4

The y-intercept equation is ...

  b = y - mx

  b = 2 -4(4) = -14

The equation of the line is y = 4x -14.

7 0
2 years ago
Help Help! Now help! ASAP!I’ll make you brainly now please
hammer [34]

Answer:

re08uwahrewrai098nyryvrhheuaviufvsdsavad <<< Don't do this please

42.3 meters

3 0
3 years ago
Use the given information to find the exact value of the trigonometric function
eimsori [14]
\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}

Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

The answer is none of the choices

7 0
1 year ago
Complete:3/4 times 20 equal 5/6 times what?
vovikov84 [41]
3/4 × 20 = 5/6y
60/4 = 5/6y
360 = 30y
y = 12
5 0
3 years ago
18(−8c+16)−13(6+3c) =
slega [8]

your final answer should be.... -3 • (61c - 70)


8 0
3 years ago
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