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chubhunter [2.5K]
3 years ago
13

1. Find the LCM of 35,14 using prime factorization.

Mathematics
1 answer:
vekshin13 years ago
3 0

Answer:

Step-by-step explanation:

1) 35 = 5 * 7

  14 = 2 * 7

LCM = 2 * 5 * 7 = 70

2)  15 = 3 * 5

    12 = 2 * 2 * 3

LCM = 2* 2 * 3 * 5 = 60

3) 5 = 5

    9 = 3 * 3

  15 = 3 * 5

LCM = 3*3 * 5 = 45

4)8 = 2 * 2 * 2 = 2³

  10 = 2 *5

   12 = 2 * 2 *3 = 2² * 3

LCM = 2³ * 3 * 5 = 120

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angle j = 90

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Step-by-step explanation:

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A theater seats 1,860 people. The last 6 shows have been sold out. Estimate the total number of people attending the last 6 show
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4 years ago
Let X denote the proportion of employees at a large firm who will choose to be covered under the firm’s medical plan, and let Y
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Answer:

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4 years ago
TRUE/FALSE. the lengths of the altitude of a hypotenuse is the geometric mean of the lengths of the segments of each of the legs
vitfil [10]

In Right Angled Triangle the lengths of the altitude of a hypotenuse is the geometric mean of the lengths of the segments of each of the legs is TRUE

What is a Right angled triangle?

A triangle is said to be right-angled if one of its angles is exactly 90 degrees. The total of the other two angles is 90 degrees. Perpendicular and the triangle's base are the sides that make up the right angle. The longest of the three sides, the third side is known as the hypotenuse.

The right triangle's three sides are interconnected. The Pythagorean Theorem explains this connection. This theorem states that the Hypotenuse2 = Perpendicular2 + Base2.

  • The right angle, or 90°, is always one angle.
  • The hypotenuse is the side with the 90° angle opposite.
  • The longest side is always the hypotenuse.
  • The other two inner angles add up to 90 degrees.

Let assume a right-angled triangle ABC, right-angled at A.

Let AD be the perpendicular drawn from vertex A on hypotenuse BC, intersecting BC at D.

Now, In right-angle triangle ABC,

By using Pythagoras Theorem, we have

AB^2 +AC^2 = BC^2

Now, In right-angle triangle ABD

Using Pythagoras Theorem, we have

AB^2 =AD^2 + BD^2

Now, In right-angle triangle ACD,

AC^2 = AD^2+CD^2

On adding equation (2) and (3), we get

AB^2 +AC^2 = 2AD^2+ BD^2 +CD^2

Now, using equation (1), the above equation can be rewritten as

2AD^2 +CD^2+BD^2 = BC^2

can be further rewritten as

2AD^2 +BD^2 +CD^2 = (CD+BD)^2\\\\BD^2 +CD^2 +2(BD)(CD)=2AD^2 +BD^2 +CD^2\\\\(BD)(CD)=AD^2

AD is geometric mean of BD and CD

Learn more about Right angled triangle from the link below

brainly.com/question/3770177

#SPJ4

8 0
2 years ago
Z=8+6x-px<br><br>solve for x
ohaa [14]
      z = 8 + 6x - px   Subtract 8 from both sides
z - 8 = 6x - px          Factor the x out of 6x - px
z - 8 = x (6 - p)         Divide both sides by (6 - p)
\frac{z - 8}{6 - p} = x   Switch the sides to make it easier to read
x = <span>\frac{z - 8}{6 - p} </span>
4 0
4 years ago
Read 2 more answers
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