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Fed [463]
3 years ago
7

The endpoints of a segment are (–5, –3) and (0, 2). What are the endpoints of the segment after it has been translated 4 units u

p? A. (–5, –7), (0, –2) b. (–5, 1), (0, 6) c. (–1, –3), (4, 2) d. (–1, 1), (4, 6)
Mathematics
2 answers:
Artist 52 [7]3 years ago
6 0

4 units up means 4 is added to the y-coordinate

  (-5, -3 + 4) and (0, 2 + 4)

= (-5, 1)         and (0, 6)

Answer: B

kiruha [24]3 years ago
5 0

Answer:

b. (-5,1) and (0,6)

Step-by-step explanation:

We have been given the endpoints of a segment. We are asked to find the coordinates of endpoints of segment after it has been translated 4 units up.

We know that when a point is translated upwards by 'a' units, we need to add 'a' units to its y-coordinate.

To find the coordinates of segment after a translation of 4 units upwards, we will add 4 to y-coordinates of both points as:

(-5,-3)\Rightarrow{-5,(-3+4)=(-5,1)

(0,2)\Rightarrow{0,(2+4)=(0,6)

Therefore, the endpoints of the segment after translation would be (-5,1) and (0,6).

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Ksju [112]
ANSWER

a = 1,2,3,6

EXPLANATION

The given equation is

ax = 6

Divide both sides by a

\Rightarrow \: \frac{ax}{a} = \frac{6}{a}

Cancel out common factors on the left hand side,

\Rightarrow \: x = \frac{6}{a}

For this solution to be a whole number, thena should be a factor of
6.

The reason is that, factors of 6 will divide exactly in to 6 without a remainder, making the answer a whole number.

The whole numbers that are factors of 6 are,

1,2,3 \: and \: 6

Note that, the set of whole numbers are,

{0,1,2,3,4,...}
3 0
3 years ago
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Answer:

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Step-by-step explanation:

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2 years ago
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balu736 [363]

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8 0
2 years ago
What is the other square root of 119 + 120i?
MakcuM [25]

Answer:

\sqrt{119+120 i}=\pm (12.24+i 4.90)

Step-by-step explanation:

Given

z = 119 + 120 i

Let \sqrt{119+120 i}=p+iq

Squaring both sides

119+120 i=p^2-q^2+2ipq

Comparing real and imaginary part

Re(LHS)=Re(RHS)

119=p^2-q^2...........................(1)

comparing Im(LHS)=Im(RHS)

120=2pq

q=\frac{60}{p}

Substitute q in 1

119=p^2-(\frac{60}{p})^2

p^4-119p^2-(68)^2=0

Let x=p^2

x^2-119x-4624=0

x=\dfrac{119\pm \sqrt{119^2+4\times 4624}}{2}

x=\frac{119\pm 180.71}{2}

we take only Positive value because p^2=x

x=149.85  

p^2=149.85

thus p=\pm 12.24

q=\pm 4.90

thus,

\sqrt{119+120 i}=\pm (12.24+i 4.90)

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2 years ago
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Answer:

Your answer is 163⁄200Step-by-step explanation:

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