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qaws [65]
3 years ago
12

Is ab tangent to the circle​

Mathematics
1 answer:
kvasek [131]3 years ago
5 0

Answer:

Yes. A tangent hits the side of a circle but does not go inside it.

Step-by-step explanation:

Please mark brainliest

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I NEED HELP PLEASE!!!! Given a real world problem, can you find the unit rate?
Yuliya22 [10]

Here is an example:

Josh bought 7 apples for $21.

21/7 is 3, so one apple costs $3. This is the unit rate.

21/7 simplifies to 3/1, that’s another way to think about it. Simplify the fraction until the denominator is 1.

Hope this helps! Tell me if you have any questions!

3 0
3 years ago
What are the rise and run of the line whose equation is y=15x?
Blababa [14]

Answer:

Rise = 15.

Run = 1.

Step-by-step explanation:

Hope this helps!

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5 0
2 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
3 years ago
Solve the equation. Round the solution to two decimal places.
navik [9.2K]
Subtract the y values to one side then the numbers to the other so as x values to one side then the numbers to the other
5 0
2 years ago
Read 2 more answers
PLEASE HELP (32 points)
baherus [9]

Answer:

B. 246 in^2

Step-by-step explanation:

A = pi r^2 theta /360   if theta is in degrees

A = * pi *(16)^2 * (110/360)

A =  *pi* 256 * 11/36

A = 2816/36 *pi

A = 245.74236 in ^2

6 0
3 years ago
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