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Vesna [10]
3 years ago
5

Imagine that each of these 40 students then made their choices from the list of four personality

Mathematics
1 answer:
Zigmanuir [339]3 years ago
8 0
To be honest my example for this is 160
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Timothy gets a phone call from some of his friends. They say they will be at the library which is 5/8 mi away. He walks at a ste
artcher [175]
I think the is answer 5/4 
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3 years ago
Help me please this is due tommorow morning
nataly862011 [7]

Answer:

E: (2, -1)

F: (3, -1)

G: (3, -2)

H: (2, -2)

Step-by-step explanation:

Reflect all coordinates over an x-axis

Because you are reflection 180 degrees over the x axis, the y-axis is going to stay the same (because you are not moving the square left or right. When reflecting over axis always add the same amount you took away from the other side, if that makes any sense.

7 0
3 years ago
How do you find the answer of -22-x=5+6x+9???
postnew [5]
-22-x=5+6x+9
-6x-x=22+5+9
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5 0
3 years ago
ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl
melamori03 [73]

Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).

Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x}  \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

\frac{AQ}{QP}=\frac{AB}{CP}  \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19

\frac{BQ}{CQ}=\frac{AB}{CP}  \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

7 0
3 years ago
what happens to the line if you change the slope a)if y = 2x + 3 and change it to y = 3x + 3 b)if you change the 2 to -2
Sidana [21]
I am pretty shur answer is a
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