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Mekhanik [1.2K]
3 years ago
6

The house numbers on the north side of Flynn Street are even. In one block, the house numbers begin 1022, 1032, 1042, and 1052.

If Taylor is at 1022 Flynn Street, how many houses away is 1082?
Mathematics
1 answer:
never [62]3 years ago
4 0

Answer:

  6 houses away

Step-by-step explanation:

You can write down the house numbers and count them:

  1032, 1042, 1052, 1062, 1072, 1082

1082 is the 6th house from where Taylor is.

___

You can also recognize that (1082 -1022)/10 = 60/10 = 6 is the number of houses away from the house at 1022.

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Vlad1618 [11]
He bought 45 bars of chocolate.

1/5 x 45 = 9 
He ate 9 bars on Tuesday.

45 - 9 = 36
He had 36 bars left.

1/12 x 36 = 3
He ate 3 bars on Wednesday

36 - 3 = 33
He had 33 bars left.

------------------------------------------------
Answer: He had 33 bars left.
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3 years ago
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Jimmy ran at a speed of m miles. how far did he run in t minutes
beks73 [17]

Answer:

(mxt) miles

Step-by-step explanation:

speed = m \: miles \: per \:minute \\ time \:  = t \: minutes \\ speed =  \frac{distance}{time}  \\ m =  \frac{d}{t}  \\ m \times t \:  = d \\ d = (m \times t)miles

hope that this is helpful.

8 0
2 years ago
Match each set of vertices with the type of triangle they form.
Andrew [12]

Answer:  The calculations are done below.


Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

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Answer:

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