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Novay_Z [31]
2 years ago
7

1. Use the data points about the distance and elevations of the camps to graph the data points where the x-coordinate is the tot

al distance traveled from the base camp and the y-coordinate is the elevation. Use the grid below or include a screenshot of the data plotted from a calculator.
Can someone please help/explain! I'm super confused and bad with word problems! The info is shown below!

Mathematics
1 answer:
Vadim26 [7]2 years ago
6 0

You should plot the total distance traveled from the base camp on the x-coordinate while the elevation should be plotted on the y-coordinate.

<h3>What is a graph?</h3>

A graph simply refers to a type of chart which is commonly used to graphically represent data on both the vertical and horizontal lines of a cartesian coordinate (x-coordinate and y-coordinate).

<h3>How to plot this graph?</h3>

In this scenario, you would plot the total distance traveled from the base camp on the x-coordinate while the elevation of the camps would be plotted on the y-coordinate as shown in the image attached below.

Read more on graphs here: brainly.com/question/25799000

#SPJ1

You might be interested in
Compare a number if it is decreased by 60% than increased by 80%.
gregori [183]

Answer:

Step-by-step explanation:

x decreased by 60% = (1-0.60)x = 0.40x

x increased by 80% = (1+0.80)x = 1.80x

8 0
2 years ago
Find all the missing sides or angles in each right triangles
astra-53 [7]
In previous lessons, we used the parallel postulate to learn new theorems that enabled us to solve a variety of problems about parallel lines:

Parallel Postulate: Given: line l and a point P not on l. There is exactly one line through P that is parallel to l.

In this lesson we extend these results to learn about special line segments within triangles. For example, the following triangle contains such a configuration:

Triangle <span>△XYZ</span> is cut by <span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span> where A and B are midpoints of sides <span><span>XZ</span><span>¯¯¯¯¯¯¯¯</span></span> and <span><span>YZ</span><span>¯¯¯¯¯¯¯</span></span> respectively. <span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span> is called a midsegment of <span>△XYZ</span>. Note that <span>△XYZ</span> has other midsegments in addition to <span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span>. Can you see where they are in the figure above?

If we construct the midpoint of side <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span> at point C and construct <span><span>CA</span><span>¯¯¯¯¯¯¯¯</span></span> and <span><span>CB</span><span>¯¯¯¯¯¯¯¯</span></span> respectively, we have the following figure and see that segments <span><span>CA</span><span>¯¯¯¯¯¯¯¯</span></span> and <span><span>CB</span><span>¯¯¯¯¯¯¯¯</span></span> are midsegments of <span>△XYZ</span>.

In this lesson we will investigate properties of these segments and solve a variety of problems.

Properties of midsegments within triangles

We start with a theorem that we will use to solve problems that involve midsegments of triangles.

Midsegment Theorem: The segment that joins the midpoints of a pair of sides of a triangle is:

<span>parallel to the third side. half as long as the third side. </span>

Proof of 1. We need to show that a midsegment is parallel to the third side. We will do this using the Parallel Postulate.

Consider the following triangle <span>△XYZ</span>. Construct the midpoint A of side <span><span>XZ</span><span>¯¯¯¯¯¯¯¯</span></span>.

By the Parallel Postulate, there is exactly one line though A that is parallel to side <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span>. Let’s say that it intersects side <span><span>YZ</span><span>¯¯¯¯¯¯¯</span></span> at point B. We will show that B must be the midpoint of <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span> and then we can conclude that <span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span> is a midsegment of the triangle and is parallel to <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span>.

We must show that the line through A and parallel to side <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span> will intersect side <span><span>YZ</span><span>¯¯¯¯¯¯¯</span></span> at its midpoint. If a parallel line cuts off congruent segments on one transversal, then it cuts off congruent segments on every transversal. This ensures that point B is the midpoint of side <span><span>YZ</span><span>¯¯¯¯¯¯¯</span></span>.

Since <span><span><span>XA</span><span>¯¯¯¯¯¯¯¯</span></span>≅<span><span>AZ</span><span>¯¯¯¯¯¯¯</span></span></span>, we have <span><span><span>BZ</span><span>¯¯¯¯¯¯¯</span></span>≅<span><span>BY</span><span>¯¯¯¯¯¯¯¯</span></span></span>. Hence, by the definition of midpoint, point B is the midpoint of side <span><span>YZ</span><span>¯¯¯¯¯¯¯</span></span>. <span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span> is a midsegment of the triangle and is also parallel to <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span>.

Proof of 2. We must show that <span>AB=<span>12</span>XY</span>.

In <span>△XYZ</span>, construct the midpoint of side <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span> at point C and midsegments <span><span>CA</span><span>¯¯¯¯¯¯¯¯</span></span> and <span><span>CB</span><span>¯¯¯¯¯¯¯¯</span></span> as follows:

First note that <span><span><span>CB</span><span>¯¯¯¯¯¯¯¯</span></span>∥<span><span>XZ</span><span>¯¯¯¯¯¯¯¯</span></span></span> by part one of the theorem. Since <span><span><span>CB</span><span>¯¯¯¯¯¯¯¯</span></span>∥<span><span>XZ</span><span>¯¯¯¯¯¯¯¯</span></span></span> and <span><span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span>∥<span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span></span>, then <span>∠<span>XAC</span>≅∠<span>BCA</span></span> and <span>∠<span>CAB</span>≅∠<span>ACX</span></span> since alternate interior angles are congruent. In addition, <span><span><span>AC</span><span>¯¯¯¯¯¯¯¯</span></span>≅<span><span>CA</span><span>¯¯¯¯¯¯¯¯</span></span></span>.

Hence, <span>△<span>AXC</span>≅△<span>CBA</span></span> by The ASA Congruence Postulate. <span><span><span>AB</span><span>¯¯¯¯¯¯¯¯</span></span>≅<span><span>XC</span><span>¯¯¯¯¯¯¯¯</span></span></span> since corresponding parts of congruent triangles are congruent. Since C is the midpoint of <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span>, we have <span>XC=CY</span> and <span>XY=XC+CY=XC+XC=2AB</span> by segment addition and substitution.

So, <span>2AB=XY</span> and <span>AB=<span>12</span>XY</span>. ⧫

Example 1

Use the Midsegment Theorem to solve for the lengths of the midsegments given in the following figure.

M, N and O are midpoints of the sides of the triangle with lengths as indicated. Use the Midsegment Theorem to find

<span><span> A. <span>MN</span>. </span><span> B. The perimeter of the triangle <span>△XYZ</span>. </span></span><span><span> A. Since O is a midpoint, we have <span>XO=5</span> and <span>XY=10</span>. By the theorem, we must have <span>MN=5</span>. </span><span> B. By the Midsegment Theorem, <span>OM=3</span> implies that <span>ZY=6</span>; similarly, <span>XZ=8</span>, and <span>XY=10</span>. Hence, the perimeter is <span>6+8+10=24.</span> </span></span>

We can also examine triangles where one or more of the sides are unknown.

Example 2

<span>Use the Midsegment Theorem to find the value of x in the following triangle having lengths as indicated and midsegment</span> <span><span>XY</span><span>¯¯¯¯¯¯¯¯</span></span>.

By the Midsegment Theorem we have <span>2x−6=<span>12</span>(18)</span>. Solving for x, we have <span>x=<span>152</span></span>.

<span> Lesson Summary </span>
8 0
3 years ago
Pls help it's fractions​
Pavlova-9 [17]

Answer:

Decimal: 0.875

Fraction: 875/1000

Step-by-step explanation:

As with our earlier problem, it turns out that we can reduce this fraction to lowest terms by dividing its numerator and denominator by 125. Doing so, we find that 0.875 = 875/1000 is equivalent to 7/8

Hope this helps!!!! :)

5 0
3 years ago
Read 2 more answers
Ram has three times as many Rupees to coins as he has rupees 5 coins if he has in all a sum of rupees 77 how many coins of Rupee
BabaBlast [244]

9514 1404 393

Answer:

  21 coins of ₹2

Step-by-step explanation:

Let x represent the number of ₹2 coins, and y the number of ₹5 coins. Then the total value of the coins is ...

  2x +5y = 77

and the relationship between numbers of coins is ...

  x = 3y

Substituting for x, we have ...

  2(3y) +5y = 77

  y = 77/11 = 7 . . . . simplify, divide by the coefficient of y

  x = 3(7) = 21 . . . . find x from the second equation

Ram has 21 of the ₹2 coins.

8 0
3 years ago
Contains the point (-1, 2) and is parallel to<br> x – 2y = -3
ivanzaharov [21]

Answer:

see explanation

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Rearrange x - 2y = - 3 into this form

Subtract x from both sides

- 2y = - x - 3 ( divide all terms by - 2 )

y = \frac{1}{2} x + \frac{3}{2} ← in slope- intercept form

with m = \frac{1}{2}

• Parallel lines have equal slopes, thus

y = \frac{1}{2} x + c ← is the partial equation

To find c substitute (- 1, 2) into the partial equation

2 = - \frac{1}{2} + c ⇒ c = 2 + \frac{1}{2} = \frac{5}{2}

y = \frac{1}{2} x + \frac{5}{2} ← in slope- intercept form

Multiply through by 2

2y = x + 5 ( subtract 2y from both sides )

0 = x - 2y + 5 ( subtract 5 from both sides )

- 5 = x - 2y, thus

x - 2y = - 5 ← in standard form

4 0
3 years ago
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