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Snowcat [4.5K]
4 years ago
11

Write the prime factorization of each number in e-g. e. 25 f. 100g. 16​

Mathematics
1 answer:
kipiarov [429]4 years ago
8 0

Answers:

25 = 5*5 which is the same as 25 = 5^2

100 = 2*2*5*5 which is the same as 100 = 2^2*5^2

16 = 2*2*2*2 which is the same as 16 = 2^4

====================================================

Explanation:

To find the prime factorization of 25, we divide by prime numbers. We see that 25/2 = 12.5 isn't a whole number, and neither is 25/3 = 8.33 (approx). But 25/5 = 5 is a whole number. The result is a prime number so we can stop here. This shows that 25 = 5*5 = 5^2

------------

We'll do the same for 100 as well. Start with dividing over 2

100/2 = 50

50/2 = 25

we can't do 25/2 as mentioned earlier, but we can say 25/5 = 5

So basically 100 = 2*2*25 = 2*2*5*5 = 2^2*5^2

-------------

The same will happen with 16 as well. We divide by 2 until we arrive at a prime number result

16/2 = 8

8/2 = 4

4/2 = 2

We have three 2's as denominators and that last result 2 is included in the set of prime factors as well to say 16 = 2*2*2*2 = 2^4

-------------

The factor tree for each is shown below. For something like 100, there are a few ways to draw out the tree. The same applies for 16 as well.

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alexandr1967 [171]

Answer:

Here's what I get.

Step-by-step explanation:

1. Rotate 180° about origin

The formula for rotation of a point (x,y) by an angle θ about the origin is  

x' = xcosθ  -  ysinθ

y' = ycosθ + xsinθ

If θ = 180°, sinθ = 0 and cosθ = -1, and the formula becomes

x' =   -x

y' =  -y

The rule is then (x, y) ⟶ (-x, -y).

H: (-3, -5) ⟶ (3, 5)

J: (-5, -3) ⟶ (5, 3)

Q: (0, -1) ⟶ (0, 1)  

The vertices of H'J'Q' are (3, 5), (5, 3), and (0, 1).

2. Rotation 90° counterclockwise about origin

cos90°  = 0 and sin90° = 1

x' = xcos90°  -  ysin90° = -y

y' = ycos90° + xsin90° =  x

The rule is then (x, y) ⟶ (-y, x).

B: (4, 5) ⟶ (-5, 4)

L: (5, 0) ⟶ (0, 5)

S: (2, 2) ⟶ (-2, 2)  

The vertices of B'L'S' are (-5, 4), (0, 5), and (-2, 2).

3. Rotation 90° clockwise about origin

cos(-90°) = 0 and sin(-90°) = -1

x' = xcos(-90°)  -  ysin(-90°) =  y

y' = ycos(-90°) + xsin(-90°) =  -x

The rule is then (x, y) ⟶ (y, -x).

F: (1, -5) ⟶ (-5, -1)

H: (-2, -3) ⟶ (-3, 2)

U: (-4, -5) ⟶ (-5, 4)  

The vertices of F'H'U' are (-5, -1), (-3, -2), and (-5, 4).

4. Rotate 180° about origin

The rule is (x, y) ⟶ (-x, -y).

J: (1, -1) ⟶ (-1, 1)

V: (2, 0) ⟶ (-2, 0)

Y: (5, -3) ⟶ (-5, 3)

R: (4, -3) ⟶ (-4, 3)  

The vertices of J'V'Y'R' are (-1, 1), (-2, 0), (-5, 3), and (-4, 3).

The figures below show your shapes before and after the rotations.

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Round 57,386 to the nearest thousand
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Part 2. State what additional information is required in order to know that the triangles are congruent by ASA.​
Helga [31]
<h3>Answer: Choice D. \angle SQR \cong \angle XQR</h3>

The red angle markers show those two angles are congruent. That's one "A" of "ASA". The S refers to a congruent pair of sides. We don't have any tickmarks to indicate congruent pairs; however, we do know that QR = QR is a shared side that overlaps (reflexive theorem). So this is the "S" in "ASA".

The thing missing is the angle Q of the top triangle, and also of the bottom triangle as well. If we know those two angles are congruent, then we have enough info to use ASA. More specifically, if we know that \angle SQR \cong \angle XQR, then we can use ASA.

One thing to notice is that the other answer choices involve side lengths and not angles. This implies that if A, B or C were one of the answers, then we would have something like SAS or SSS. But instead we want ASA. So we can immediately rule choices A,B, and C out.

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First one:

cos(A)=AC/AB=3/4.24

cos(B)=BC/AB=3/4.24

Cos(A)/cos(B)=AC/AB / (BC/AB) = AC/AB * AB/BC = AC/BC=3/3=1


Second one:

To solve this problem, we have to ASSUME AFE is a straight line, i.e. angle EFB is 90 degrees. (this is not explicitly given).

If that's the case, AE is a transversal of parallel lines AB and DE.

And Angle A is congruent to angle E (alternate interior angles).

Therefore sin(A)=sin(E)=0.5

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