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Nastasia [14]
3 years ago
10

Can someone please respond to me and help me out with these 4 problems I must show my work and include an expression , evaluatio

n and label for my answers or I won’t get full credit 20-22 questions 17-19 I just include an equation, answer and label for each question

Mathematics
1 answer:
AlladinOne [14]3 years ago
4 0

Answer: A train traveled486 miles in 7 hours

They would be going at least 40 - 70

Step-by-step explanation:

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How can you make 2 cups of chocolate milk a gallon?<br> I know that 32 = a gallon
const2013 [10]
Well you could use the equation 32=2x or 32=(2*x) but if you're looking for an answer that is other than turning this into an equation, then it would be impossible to find an answer. I hope this helped ^^
5 0
2 years ago
Kyle is practicing for a 3-mile race. His normal time is 23 minutes 25 seconds. Yesterday it took him only 21 minutes 38 seconds
diamong [38]

Answer:2:24

Step-by-step explanation:

7 0
3 years ago
What is the number of real solutions?
hichkok12 [17]
2 solutions, the root to the equation is the same as the root to the solution also known as “double roots”
8 0
2 years ago
Can someone help me with this? Please this would mean so much.
Paha777 [63]

Answers:

x = angle MLN = 35 degrees

angle FLJ = 55 degrees

=================================================

Explanation:

This problem is fairly tricky if you're not sure what to look for. A slight clue is that they've marked a red point that strangely doesn't have a label on it. It's a fairly small point but it's definitely there if you look closely. While this point's location isn't exactly what we want, we're fairly close. Start at point L and draw a ray through the center point P. Ray LP will intersect the circle at point A, as shown in the diagram below.

From here, draw segments FA and MA. Now notice that inscribed angle LMF = 55 and inscribed angle FAL both subtend the same arc. The term "subtend" basically means "cut off". This arc that the inscribed angles subtend is minor arc FL. A minor arc is where you travel the shorter path around the circle, which indicates its measure is less than 180 degrees.

Since inscribed angles LMF and FAL subtend the same minor arc, this makes the inscribed angles to be congruent.

In short: angle FAL is 55 degrees

----------------------

Segment LA goes through the center P. Through Thales theorem, we know that inscribed angle LFA is 90 degrees. Consequently, we can determine that inscribed angle FLA is 90-55 = 35 degrees.

Segment JL is tangent to the circle, meaning that angle ALJ is 90 degrees. So angle FLJ is 90-35 = 55 degrees.

It's not a coincidence that angle FLJ, angle FAL, and angle LMF are the same measure.

----------------------

We found that angle FAL was 55 degrees. Applying Thales theorem again shows that angle MAF is 90 degrees. Therefore, angle LAM is 90-55 = 35 degrees.

Focus now on triangle LMA. This is also a right triangle (Thales Theorem). The upper acute angle we found was angle LAM = 35, so the lower acute angle is ALM = 55.

Then we can find angle MLN = (angle ALN) - (angle ALM) = 90 - 55 = 35

In short, angle MLN = 35 degrees

Similar to the previous section, it is not a coincidence that angles LFM, LAM and MLN are the same measure.

------------------------

As an alternative, since we know angle FLJ = 55, and angle FLM is 90 degrees, this means...

(angle FLJ)+(angle FLM) + (angle MLN) = 180

55 + 90 + angle MLN = 180

145 + angle MLN = 180

angle MLN = 180 - 145

angle MLN = 35 degrees

7 0
2 years ago
Runner A is initially 6.0 km west of a flagpole and is running with a constant velocity of 3.0 km/h due east. Runner B is initia
ArbitrLikvidat [17]

Answer:

The distance of the two runners from the flagpole is 0.0882 km

Step-by-step explanation:

We are going to use the minus sign when the runner is on the west and the plus sign when the runner is on the east.

Then, Taking into account  the runner A is 6 km west and is running with a constant velocity of 3 Km/h east, the distance of the runner A from the flagpole is given by the following equation:

Xa = -6 Km + (3 Km/h)* t

Where Xa is the position of the runner A from the flagpole and t is the time in hours.

At the same way the distance of the runner B, Xb, from the flagpole is given by the following equation:

Xb = 7.4 Km - (3.8 Km/h)*t

Then, the two runners cross their path when Xa is equal to Xb, so if we solve this equation for t, we get:

              Xa = Xb

     -6 + (3*t) =  7.4 - (3.8*t)

(3.8*t) + (3*t) = 7.4 + 6

         (6.8*t) = 13.4

                  t = 13.4/6.8

                  t = 1.9706

Therefore, at time t equal to 1.9706 hours, both runners cross their path. The distance of the two runners from the flagpole can be calculated replacing the value of t in equation for Xa or in equation for Xb as:

Xa = -6 Km + (3 Km/h)* t

Xa = -6 Km + (3 Km/h)*(1.9706 h)

Xa = -0.0882 Km

That means that both runners are 0.0882 Km west of a flagpole.

4 0
2 years ago
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