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Luden [163]
3 years ago
6

Solve the equation 6=y-5 which step can you use to solve the equation 6+y-5

Mathematics
1 answer:
bazaltina [42]3 years ago
3 0

Answer:

1.)6=y-5

On transposing y to L.H.S, we obtain

6+5=y

11=y

Step-by-step explanation:

Hope this helps you

If yes then do mark my answer as brainliest

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Emma has a spool of ribbon that has 5 meters on it. She cut a length of ribbon that was 0.9 meters long. Then she cut another le
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Answer:

2.6 meters.

Step-by-step explanation:

5-0.9 is 4.1, minus 1.5 is 2.1.

7 0
3 years ago
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How do I solve this problem?
Liono4ka [1.6K]

Answer:

10100

Step-by-step explanation:

Use this formula,

\frac{n}{2} (2 {a} + d(n - 1)

where n is the amount of even intergers, a is the starting term, d is the common difference.

  • the amount of even intergers from 2 to 200 is 100
  • The starting number is 2
  • The common difference is 2

\frac{100}{2} (4 + 2(99)

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50(202)

10100

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2 years ago
Which of the following points lies on the line whose equation is y = 2x - 3?
mariarad [96]
It's supposed to be (0,-3) but I guess your answer should be A
6 0
3 years ago
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8. In right triangle LMN shown below, altitude MK is drawn to LN from M. IF LM = 12 and LK =10, then
Elena-2011 [213]

The length of the KN is 4.4

Step-by-step explanation:

We know from Pythagoras theorem  

In a right angle ΔLMN  

Base² + perpendicular² = hypotenuse ²

From the properties of triangle we also know that altitudes are ⊥ on the sides they fall.

Hence ∠LKM = ∠NKM = 90 °

Given values-

LM=12

LK=10

Let KN be “s”

⇒LN= LK + KN

⇒LN= 10+x        eq 1

Coming to the Δ LKM

⇒LK²+MK²= LM²

⇒MK²= 12²-10²

⇒MK²= 44           eq 2

Now in Δ MKN

⇒MK²+ KN²= MN²

⇒44+s²= MN²       eq 3

In Δ LMN

⇒LM²+MN²= LN²

Using the values of MN² and LN² from the previous equations

⇒12² + 44+s²= (10+s) ²

⇒144+44+s²= 100+s²+20s

⇒188+s²= 100+s²+20s    cancelling the common term “s²”

⇒20s= 188-100

∴ s= 4.4

Hence the value of KN is 4.4  

8 0
3 years ago
Find the volume of a hemisphere whose radius is 7.5
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The answer is 1767.16 ;)
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