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Mazyrski [523]
1 year ago
9

All the students in the sixth grade either purchased their lunch or brought their lunch from home on Monday 24% of the students

purchased their lunch 190 students brought their lunch from home how many students ante in the sixth grade?
Mathematics
1 answer:
Alla [95]1 year ago
5 0

I believe it is 60 but I’m not sure

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The image shows parallel lines cut by a transversal. The expressions represent unknown angle measurements. What is the value of
Anna35 [415]

Answer:

x = 8

Step-by-step explanation:

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1 year ago
a rectangle rug has a perimeter of 146 ft the width of the rug is 5 feet more than three times the length find the length and th
dem82 [27]

Answer:

The length = 56 feet and the width = 17 feet.

Step-by-step explanation:

We can set up 2 equations to solve this. Let the length of the rug be x, then

x = 3w + 5    where w = the width.   ( looks like you got the width and the length mixed up. The length is the longest side)

The perimeter = 2x + 2w = 146 so we have the 2 equations:

x = 3w + 5

2x + 2w = 146

Now we substitute for x in the second equation:

2(3w + 5) + 2w = 146

6w + 10 + 2w = 146

8w = 136

w = 17 feet,

and x = 3(17) + 5 =  56 feet.

7 0
1 year ago
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Stephen says that the numbers 38 and 40 are relatively prime. explain why he is incorrect in making this statement
bonufazy [111]

When you say that two numbers are relatively prime, it means that both numbers do not have a common factor except 1. Stephen is incorrect because 38 and 40 are both multiples of 2.

I hope I answered your question. Thank you <span>J</span>

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2 years ago
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HELP!!<br> What is the equation of the circle shown in the graph?
ale4655 [162]

9514 1404 393

Answer:

  (x +6)^2 +(y -4)^2 = 36

Step-by-step explanation:

The center is (-6, 4) and the radius is 6. Putting those into the standard form equation, you have ...

  (x -h)^2 +(y -k)^2 = r^2 . . . . . . center (h, k), radius r

  (x -(-6))^2 +(y -4)^2 = 6^2 . . . . numbers filled in

  (x +6)^2 +(y -4)^2 = 36 . . . . . . cleaned up a bit

8 0
1 year ago
Given: The coordinates of triangle PQR are P(0, 0), Q(2a, 0), and R(2b, 2c).
masha68 [24]

Answer:

The line containing the midpoints of two sides of a triangle is parallel to the third side ⇒ proved down

Step-by-step explanation:

* Lets revise the rules of the midpoint and the slope to prove the

  problem

- The slope of a line whose endpoints are (x1 , y1) and (x2 , y2) is

  m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

- The mid-point of a line whose endpoints are (x1 , y1) and (x2 , y2) is

  (\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

* Lets solve the problem

- PQR is a triangle of vertices P (0 , 0) , Q (2a , 0) , R (2b , 2c)

- Lets find the mid-poits of PQ called A

∵ Point P is (x1 , y1) and point Q is (x2 , y2)

∴ x1 = 0 , x2 = 2a and y1 = 0 , y2 = 0

∵ A is the mid-point of PQ

∴ A=(\frac{0+2a}{2},\frac{0+0}{2})=(\frac{2a}{2},\frac{0}{2})=(a,0)

- Lets find the mid-poits of PR which called B

∵ Point P is (x1 , y1) and point R is (x2 , y2)

∴ x1 = 0 , x2 = 2b and y1 = 0 , y2 = 2c

∵ B is the mid-point of PR

∴ B=(\frac{0+2b}{2},\frac{0+2c}{2})=(\frac{2b}{2},\frac{2c}{2})=(b,c)

- The parallel line have equal slopes, so lets find the slopes of AB and

  QR to prove that they have same slopes then they are parallel

# Slope of AB

∵ Point A is (x1 , y1) and point B is (x2 , y2)

∵ Point A = (a , 0) and point B = (b , c)

∴ x1 = a , x2 = b and y1 = 0 and y2 = c

∴ The slope of AB is m=\frac{c-0}{b-a}=\frac{c}{b-a}

# Slope of QR

∵ Point Q is (x1 , y1) and point R is (x2 , y2)

∵ Point Q = (2a , 0) and point R = (2b , 2c)

∴ x1 = 2a , x2 = 2b and y1 = 0 and y2 = 2c

∴ The slope of AB is m=\frac{2c-0}{2b-2a}=\frac{2c}{2(b-c)}=\frac{c}{b-a}

∵ The slopes of AB and QR are equal

∴ AB // QR

∵ AB is the line containing the midpoints of PQ and PR of Δ PQR

∵ QR is the third side of the triangle

∴ The line containing the midpoints of two sides of a triangle is parallel

  to the third side

6 0
1 year ago
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