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WITCHER [35]
3 years ago
5

What is the rate of change of the line

Mathematics
2 answers:
jolli1 [7]3 years ago
8 0

Answer:

5 feet per second

Step-by-step explanation:


butalik [34]3 years ago
3 0
B 5 feet per second
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The side of an Equileteral triangle is 12cm. What is its Area?
elena55 [62]

Answer:

A = 62.35 cm²

Step-by-step explanation:

Use the area formula A = \frac{\sqrt{3}a^2}{4}, where a is the side length.

Plug in the values:

A = \frac{\sqrt{3}(12^2)}{4}

A = \frac{\sqrt{3}(144)}{4}

A = 62.35 cm²

3 0
4 years ago
Please help whoever answers this correctly I'll mark your answer brainliest​
djyliett [7]

Answer: The 2nd and 3rd one are correct

Step-by-step explanation:

After reflecting it, the square itself would not change, only the position of the square. Therefore, the same line segments from the beginning which were given would be parallel.

6 0
2 years ago
Read 2 more answers
Estimated m 60=4m <br><br><br><br><br> 4x0=
11111nata11111 [884]

[ Answer ]


M = <em><u>15</u></em>

4 * 0 = <u><em>0</em></u>

[ Explanation ]


60 / 4 = 15

15 * 4 = 60



4 * 0 = 0


<> Eclipsed <>


4 0
3 years ago
Six sophomores and 14 freshmen are competing for two alternate positions on the debate team. Which expression represents the pro
Elanso [62]

Answer:

<em>Choose the first alternative</em>

\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}

Step-by-step explanation:

<u>Probabilities</u>

The requested probability can be computed as the ratio between the number of ways to choose two sophomores in alternate positions (N_s) and the total number of possible choices (N_t), i.e.

\displaystyle P=\frac{N_s}{N_t}

There are 6 sophomores and 14 freshmen to choose from each separate set. There are 20 students in total

We'll assume the positions of the selections are NOT significative, i.e. student A/student B is the same as student B/student A.

To choose 2 sophomores out of the 6 available, the first position has 6 elements to choose from, the second has now only 5

_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C} \text{ ways to do it}

The total number of possible choices is

_{2}^{20}\textrm{C} \text{ ways to do it}

The probability is then

\boxed{\displaystyle P=\frac{_{1}^{6}\textrm{C}\ _{1}^{5}\textrm{C}}{_{2}^{20}\textrm{C}}}

Choose the first alternative

7 0
3 years ago
Read 2 more answers
I need help figuring which of the answers are it I need to pick 3 please help
Firdavs [7]
A , b & c ?
lmk if i’m wrong tho
7 0
3 years ago
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