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const2013 [10]
3 years ago
6

If M(x, y) is the midpoint of AB with A(3, 5) and B(-4, 8), then which Quadrant is M located in? (Points : 1) Quadrant I Quadran

t II Quadrant III Quadrant IV Question 2. 2. Given A(9, 11), B(3, -4), C(6, -9), and D(-7, 7), then the midpoint of which line located in Quadrant II? (Points : 1) AB BC AC BD Question 3. 3. Given A(6, -17), B(-6, 17), C(4, 9), D(-4, 4), the midpoint of which segment is at the origin? (Points : 1) AB BC AC AD
Mathematics
1 answer:
Alex Ar [27]3 years ago
8 0
M is located in Quadrant II, the line BD has a midpoint located in Quadrant II, AD has a midpoint at the origin.
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Answer:

Liam can paint 18 bird houses.

Step-by-step explanation:

9/10 divided by 1/20 = 18

8 0
2 years ago
Find 8 and one third percent of 144
madam [21]
\bf 8\frac{1}{3}\implies \cfrac{8\cdot 3+1}{3}\implies \cfrac{25}{3}
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now\qquad \cfrac{\frac{25}{3}}{100}\implies \cfrac{\frac{25}{3}}{\frac{100}{1}}\implies \cfrac{25}{3}\cdot \cfrac{1}{100}\implies \cfrac{1}{3\cdot 4}\implies \cfrac{1}{12}
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8\frac{1}{3}\%\ of\ 144\implies  \left( \cfrac{\frac{25}{3}}{100} \right)\cdot 144\implies \cfrac{1}{12}\cdot 144\implies \cfrac{144}{12}\implies 12
3 0
2 years ago
HELP 5 STAR + THANKS What series of transformations from △ABC to ​ △DEF ​ shows that △ABC≅△DEF ? a reflection across the y-axis
nirvana33 [79]
In the given triangle, the verteces are A(-4, 1), B(-6, 5), C(-1, 2).
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A translation of 1 unit to the right will result in A"(-3, -1), B"(-5, -5), C"(0, -2)
A translation of 1 unit down will result in A"'(-3, -2), B"'(-5, -6), C"'(0, -3) which corresponds to points DEF.
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7 0
2 years ago
The width of a rectangle is shown below: A coordinate plane with a point A at negative 3, 3 and C at negative 3, negative 2. If
noname [10]
The correct answer should be A, at positive 3 , and D , at positive 3
5 0
3 years ago
Read 2 more answers
Please help me with the work because i dont understand this.
Alexandra [31]

Answer:

The dimension are 20 ft by 15 ft

Step-by-step explanation:

A = l*w

P = 2(l+w)

we know the area = 300

and the perimeter = 70

300 = lw

70 = 2 (l+w)

divide by 2

70/2 = 2/2 (l+w)

35 = l+w

subtract w

35-w = l+w-w

35 -w =l

substitute this into 300 = lw

300 = (35-w) * w

distribute

300 = 35w - w^2

subtract 300 from each side

0 = -w^2 +35w - 300

divide by -1

0 = w^2 - 35w + 300

factor

0= (w-15) (w-20)

using the zero product property

w-15 = 0   w-20 =0

so w=15, w=20

if w=15   then l=35 -w  l = 35-15   l=20

if w=20  then l=35-w   l= 35-20 = 15


The dimension are 20 ft by 15 ft

7 0
3 years ago
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