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gizmo_the_mogwai [7]
2 years ago
14

What is the length of Line segment B C?

Mathematics
2 answers:
SashulF [63]2 years ago
5 0

Answer:

We are given that the length of side BC is congruent to AB so they have an equal measure. Based on this we can make an equation where their lengths are equal, and then solve for x:

2x - 6 = x + 17

x - 6 = 17

x = 23

Now we can plug x into the expression 2x - 6:

BC = 2(23) - 6

BC = 46 - 6

BC = 40

Step-by-step explanation:

Katen [24]2 years ago
3 0
23 because ofc was like a joke and he was never mad
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If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t
sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, 6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296

This implies that 6^{4} exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, 6^{3}=216 and 216\times 5=1080, 216\times 4=864. This implies that 216\times 5 exceeds 1052 and thus there can be at most 4 groups of 6^{3}.

Then,

1052-4\times6^{3}

1052-4\times216

1052-864

188

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, 6^{2}=36 and 36\times 5=180, 36\times 6=216. This implies that 36\times 6 exceeds 188 and thus there can be at most 5 groups of 6^{2}.

Then,

188-5\times6^{2}

188-5\times36

188-180

8

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, 6^{1}=6 and 6\times 1=6, 6\times 2=12. This implies that 6\times 2 exceeds 8 and thus there can be at most 1 group of 6^{1}.

Then,

8-1\times6^{1}

8-1\times6

8-6

2

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, 6^{0}=1 and 1\times 2=2. This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 (6^{3}), 5 groups of second power of 6 (6^{2}), 1 group of first power of 6 (6^{1}) and 2 groups of zeroth power of 6 (6^{0}).

Then,

1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}

So, the number 1052, written as a base 6 number is 4512.

Learn more about change of base of numbers here:

brainly.com/question/14291917

6 0
2 years ago
What is the greatest common factor of 62, 41, and 71?
garik1379 [7]
62= 1 \cdot 2 \cdot 31 \\
41=1 \cdot 41 \\
71=1 \cdot 71 \\ \\
GCF(62,41,71)=1
3 0
3 years ago
Billy wants to live in the area defined by y < 3x − 6. Explain how you can identify the houses in which Billy is interested i
Nadusha1986 [10]
Check the picture below.

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and then you do a "true" or "false" check.

now, if you look at the line below, it splits the grid in two sections, so, let's check a point in either section, if one is false, the other is true and the other way around.

so, let's check the region on the left-hand-side, hmmmm say point 0,0, the origin.

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y = 0,  x = 0  thus

0 < 3(0) - 6

0 < -6   <--- now, is that really true? is 0 lesser than -6? not quite, is false.

recall that on the negative side, the closer to 0, the larger the value, so -1 is much larger than -1,000,000.

because the point 0,0 yielded a false inequality, that region is the "false region" and thus not shaded, therefore, the other side must be the "true region", and thus we shade that then.

we can run a quick check on that btw, say on point 4,4

y =4,   x = 4  thus

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4 < 12 - 6

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What plus what equals 234
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4 0
3 years ago
Read 2 more answers
The circumference of circle p is 800 mm, the circumference of circle q is 200 cm, and the circumference of circle r is 4 m. What
shepuryov [24]

Let us use the following conversions:

1 cm = 10 mm

1 meter = 100 cm = 1000 mm

Given:

p = 800 mm

q = 200 cm = 2000 mm

r= 4 m = 4000

X = sum of the distance around each circle

X = p+q+r

X = 800+2000+4000

<span>X = 6800 mm or 680 cm or 6.80 m</span>

7 0
3 years ago
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