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alexdok [17]
3 years ago
14

At soccer camp Bruce spent one hour and 30 minutes practicing defense and 2 hours shooting. if the camp ended at 2:15 PM what ti

me does it start?
Mathematics
2 answers:
Likurg_2 [28]3 years ago
7 0
2:15 - 2 hours would = 12:15 subtract it by 1 hour more it would be: 11:15 - 30 minutes would make it: 10:45 so the camp started at 10:45
Sati [7]3 years ago
4 0
2:15-2 hrs is 12:15 - 1 hr is 11:15-30 min is 10:45... There is your answer!
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Is 2 units by 7 units and 11 units and 37 units a scale copy
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Answer:

what are u asking

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3 years ago
Add the two expressions -4x + 3 and 2x - 6 enter your answer in the Box PLZ HELP FASt
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-4x+3+2x-6
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4 years ago
How to solve this trig
n200080 [17]

Hi there!

To find the Trigonometric Equation, we have to isolate sin, cos, tan, etc. We are also given the interval [0,2π).

<u>F</u><u>i</u><u>r</u><u>s</u><u>t</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

What we have to do is to isolate cos first.

\displaystyle  \large{ cos \theta =  -  \frac{1}{2} }

Then find the reference angle. As we know cos(π/3) equals 1/2. Therefore π/3 is our reference angle.

Since we know that cos is negative in Q2 and Q3. We will be using π + (ref. angle) for Q3. and π - (ref. angle) for Q2.

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>2</u>

\displaystyle \large{ \pi -  \frac{ \pi}{3}  =  \frac{3 \pi}{3}  -  \frac{  \pi}{3} } \\  \displaystyle \large \boxed{ \frac{2 \pi}{3} }

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u>\displaystyle \large{ \pi  +   \frac{ \pi}{3}  =  \frac{3 \pi}{3}   +   \frac{  \pi}{3} } \\  \displaystyle \large \boxed{ \frac{4 \pi}{3} }</u>

Both values are apart of the interval. Hence,

\displaystyle \large \boxed{ \theta =  \frac{2 \pi}{3} , \frac{4 \pi}{3} }

<u>S</u><u>e</u><u>c</u><u>o</u><u>n</u><u>d</u><u> </u><u>Q</u><u>u</u><u>e</u><u>s</u><u>t</u><u>i</u><u>o</u><u>n</u>

Isolate sin(4 theta).

\displaystyle \large{sin 4 \theta =  -  \frac{1}{ \sqrt{2} } }

Rationalize the denominator.

\displaystyle \large{sin4 \theta =  -  \frac{ \sqrt{2} }{2} }

The problem here is 4 beside theta. What we are going to do is to expand the interval.

\displaystyle \large{0 \leqslant  \theta < 2 \pi}

Multiply whole by 4.

\displaystyle \large{0 \times 4 \leqslant  \theta \times 4 < 2 \pi \times 4} \\  \displaystyle \large \boxed{0 \leqslant 4 \theta < 8 \pi}

Then find the reference angle.

We know that sin(π/4) = √2/2. Hence π/4 is our reference angle.

sin is negative in Q3 and Q4. We use π + (ref. angle) for Q3 and 2π - (ref. angle for Q4.)

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>3</u>

<u>\displaystyle \large{ \pi +  \frac{ \pi}{4}  =  \frac{ 4 \pi}{4}  +  \frac{ \pi}{4} } \\  \displaystyle \large \boxed{  \frac{5 \pi}{4} }</u>

<u>F</u><u>i</u><u>n</u><u>d</u><u> </u><u>Q</u><u>4</u>

\displaystyle \large{2 \pi -  \frac{ \pi}{4}  =  \frac{8 \pi}{4}  -  \frac{ \pi}{4} } \\  \displaystyle \large \boxed{ \frac{7 \pi}{4} }

Both values are in [0,2π). However, we exceed our interval to < 8π.

We will be using these following:-

\displaystyle \large{ \theta + 2 \pi k =  \theta \:  \:  \:  \:  \:  \sf{(k  \:  \: is \:  \: integer)}}

Hence:-

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>3</u>

\displaystyle \large{ \frac{5 \pi}{4}  + 2 \pi =  \frac{13 \pi}{4} } \\  \displaystyle \large{ \frac{5 \pi}{4}  + 4\pi =  \frac{21 \pi}{4} } \\  \displaystyle \large{ \frac{5 \pi}{4}  + 6\pi =  \frac{29 \pi}{4} }

We cannot use any further k-values (or k cannot be 4 or higher) because it'd be +8π and not in the interval.

<u>F</u><u>o</u><u>r</u><u> </u><u>Q</u><u>4</u>

\displaystyle \large{ \frac{ 7 \pi}{4}  + 2 \pi =  \frac{15 \pi}{4} } \\  \displaystyle \large{ \frac{ 7 \pi}{4}  + 4 \pi =  \frac{23\pi}{4} } \\  \displaystyle \large{ \frac{ 7 \pi}{4}  + 6 \pi =  \frac{31 \pi}{4} }

Therefore:-

\displaystyle \large{4 \theta =  \frac{5 \pi}{4} , \frac{7 \pi}{4} , \frac{13\pi}{4} , \frac{21\pi}{4} , \frac{29\pi}{4}, \frac{15 \pi}{4} , \frac{23\pi}{4} , \frac{31\pi}{4}  }

Then we divide all these values by 4.

\displaystyle \large \boxed{\theta =  \frac{5 \pi}{16} , \frac{7 \pi}{16} , \frac{13\pi}{16} , \frac{21\pi}{16} , \frac{29\pi}{16}, \frac{15 \pi}{16} , \frac{23\pi}{16} , \frac{31\pi}{16}  }

Let me know if you have any questions!

3 0
3 years ago
Two points are separated on the ground by a distance of 43 km. Can they be placed on a 1 / 25000th map with dimensions 1.32 m on
Wittaler [7]

Answer:

No

Explanation:

The distance between the two points on the ground = 43km

\text{Scale}=\frac{1}{25000}

First, we calculate the distance between the two points in meters.

• 43km = 43,000 m.

Next, determine the distance it will take using the scale.

\begin{gathered} \frac{1}{25000}=\frac{x}{43000} \\ 25000x=43000 \\ x=\frac{43000}{25000} \\ x=1.72m \end{gathered}

The distance between the points using the scale is 1.72m.

Next, we see that the dimensions (1.32 m by 0.85 m) are smaller than the length.

As a final check, calculate the length of the diagonal of the map. The diagonal is usually the longest part.

\text{Diagonal}=\sqrt[]{1.32^2+0.85^2}=1.57m

Since 1.57m is still less than 1.72m, the two points cannot be placed on the map.

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1 year ago
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(-7/2) / 2 = x
-7/2 * 1/2 = x
- 7/4 (or - 1 3/4) = x
4 0
4 years ago
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