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ioda
3 years ago
7

Please help !!!!!!!!!!!!!

Mathematics
1 answer:
saw5 [17]3 years ago
8 0
Since they are opposite angles about the intersection of two straight lines they are equal angles....

5x+12=6x-10  subtract 12 from both sides

5x=6x-22  subtract 6x from both sides

-x=-22  divide both sides by -1

x=22°

So those angles are 5*22+12=122°

Now HJI is an adjacent angle so it is 180°-122°=58°
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Step-by-step explanation:

  1. 90-27=63°
  2. <2
  3. <2,<4
  4. 360°
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2 years ago
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1/2x + 3y=4 for x when x = 6
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Wouldn't it be y=1/3?
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3 years ago
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Walk fifty meters at 30o north or east from the old oak tree. (2) Turn 45o to your left (you should now be facing 75o north of e
saw5 [17]

Answer:

A straight line of approximately 75 meters, 1.4º north

Step-by-step explanation:

Hi, let's make it step by step to make it clearer

1) If we walk 50 meters in 30º angle Northeast, assuming the Old Oak tree is the point 0,0 and we're dealing with vectors in R^{2}. To say 30º Northeast is 30º clockwise (or 60º counter clockwise).

2) Then there was a the turning point to the left. If I turn to the left, on my compass 45º , I'll face 75º northeast.

3) Finally, the last vector leads to the treasure from the Old Oak Tree, i.e. the resultant.

So, let's calculate the norm which is the length of the each vector.

1) Graphing them we can find the points, then the components and then calculate the norm, the length of each vector.  

Since the Oak Tree is on (0,0). The turning point (50,86.61) and the Rock (R=(1.4,74,85) we can write the following vectors:

\vec{u}=\left \langle 50,86.61 \right \rangle\\\vec{v}=\left \langle -48.6,-11.76\right \rangle\\\vec{w}=\left \langle 1.4,74.85 \right \rangle

Now, let's calculate each vector length by calculating the norm.

\left \| \vec{u} \right \|=\sqrt{50^{2}+86.6^2}=100\\\left \| \vec{v} \right \|=\sqrt{(-48.6)^2+(-11.76)^2}=50\\\left \| \vec{u} \right \|=\sqrt{(1.4)^2+(74.85)^2}=74.86

The path is almost 75 meters. And since it is less than 15º degrees to the left of the North (or to the right) its direction is still north of the Old Oak Tree.

8 0
3 years ago
Logan used 12 centimeters of tape to wrap 4 presents. How much tape will Logan need in all if he had wrap 8 presents? Assume the
inn [45]

Answer:

24 centimeters

Step-by-step explanation:

6 0
3 years ago
Before the pandemic cancelled sports, a baseball team played home games in a stadium that holds up to 50,000 spectators. When ti
spayn [35]

Answer:

(a)D(x)=-2,500x+60,000

(b)R(x)=60,000x-2500x^2

(c) x=12

(d)Optimal ticket price: $12

Maximum Revenue:$360,000

Step-by-step explanation:

The stadium holds up to 50,000 spectators.

When ticket prices were set at $12, the average attendance was 30,000.

When the ticket prices were on sale for $10, the average attendance was 35,000.

(a)The number of people that will buy tickets when they are priced at x dollars per ticket = D(x)

Since D(x) is a linear function of the form y=mx+b, we first find the slope using the points (12,30000) and (10,35000).

\text{Slope, m}=\dfrac{30000-35000}{12-10}=-2500

Therefore, we have:

y=-2500x+b

At point (12,30000)

30000=-2500(12)+b\\b=30000+30000\\b=60000

Therefore:

D(x)=-2,500x+60,000

(b)Revenue

R(x)=x \cdot D(x) \implies R(x)=x(-2,500x+60,000)\\\\R(x)=60,000x-2500x^2

(c)To find the critical values for R(x), we take the derivative and solve by setting it equal to zero.

R(x)=60,000x-2500x^2\\R'(x)=60,000-5,000x\\60,000-5,000x=0\\60,000=5,000x\\x=12

The critical value of R(x) is x=12.

(d)If the possible range of ticket prices (in dollars) is given by the interval [1,24]

Using the closed interval method, we evaluate R(x) at x=1, 12 and 24.

R(x)=60,000x-2500x^2\\R(1)=60,000(1)-2500(1)^2=\$57,500\\R(12)=60,000(12)-2500(12)^2=\$360,000\\R(24)=60,000(24)-2500(24)^2=\$0

Therefore:

  • Optimal ticket price:$12
  • Maximum Revenue:$360,000

3 0
3 years ago
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