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Anettt [7]
1 year ago
11

18+6=__+__=__ /\ 2 4

Mathematics
2 answers:
Anna71 [15]1 year ago
5 0
18+ 6 equals 14 just add
Arturiano [62]1 year ago
3 0

Answer:

18 pluse 6 equals 14

Step-by-step explanation:

just add

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Help meee plssss!!!!!
lakkis [162]

Circumference of circle is 62.8 inches.

Step-by-step explanation:

Using Formula

Circumference of circle = 2 π r

Substitute the values

Circumference of circle = 2 × 3.14 × 10 inches.

multiply the numbers

Circumference of circle = 62.8 inches

Hence, the circumference of circle is 62.8 inches.

8 0
3 years ago
If LMN is equal or congruent to XYZ, which congruences are true by CPCTC? check all that apply.
Kamila [148]
D. Z ~ N
E. ML~ YX
F. X~ L
7 0
3 years ago
WHOEVER FINISHES FIRST I WILL MARK AS BRAINLIEST!!!
Lady bird [3.3K]

Answer:

-350%  -0.4%   -40%  -37.5% hope this helps!

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
I’m confused on this one
dedylja [7]

It's a parallelogram so if one angle is 90 degrees they all will be because of transverse angles and all that good stuff.


So we're given the diagonal of a rectangle and one side and we're asked to find the other. The diagonal of a rectangle is the hypotenuse of the right triangle whose legs are the sides of the rectangle. So this is a Pythagorean Theorem question in disguise:


AB^2 + AD^2 = BD^2


AD^2 = BD^2 - AB^2


AD = \sqrt{ BD^2 - AB^2 }


AD = \sqrt{149^2 - 51^2} = \sqrt{19400} = 140


Answer: 140 cm


I don't recall seeing this Pythagorean Triple before.





3 0
3 years ago
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
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