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Dmitry_Shevchenko [17]
3 years ago
8

Could someone help with this inequality? 20≥−3.2(c−4.3) thanks >o

Mathematics
2 answers:
IrinaK [193]3 years ago
4 0
Your solution would be -1.95<=x . btw couldn’t do the less than or equal to so i put that.
givi [52]3 years ago
4 0

Answer:

c ≥ 1.95

Step-by-step explanation:

If

20≥−3.2(c−4.3),

then,

20 ≥ −3.2c + 13.76

20 - 13.76 ≥ −3.2c

6.24 ≥ −3.2c

Divide through by -3.2 and this reverses the sign (due to the division by negative number)

c ≥ 1.95.

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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
How do i solve tan 6° = f/2.3
kondor19780726 [428]
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5 0
1 year ago
You have two cubes that you filled with water to make ice cubes. The first cube has a side length of 3 units. The second cube ha
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2 years ago
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6. Find the area of a triangle with sides a = 5, b = 8, and c = 11.​
Colt1911 [192]
<h2>20</h2>

Hi, I’m Lena and I’ll be here to answer your question with a good-quality answer. If there's something I can improve, let me know! ツ

♡ Let’s look at the main important things that will lead you in the correct path.

♥ <em>a </em>+<em> b</em> usually = <em>c!</em> Unless the book asks you to find a specific variable, you will most likely be looking for <em>c</em>.

♥ <em>A</em> = <em>bh</em>/2! This is the formula for the area of a triangle. You can also read this as A = base * height/2. The good thing about this formula is you don't need a picture. You can write it as <em>a</em> * b<em> </em>or <em>b</em> * <em>a!</em>

<em />

♡ Now, let's solve!

♥ Plug the numbers into the formula: 5*8/2.

♥ Solve! In this case, we do not need c: 40/2 = 20.

Your answer should be 20! Please do not hesitate to ask me any questions.

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Step-by-step explanation:

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