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lorasvet [3.4K]
3 years ago
8

Pls Help Choose the correct word to complete the statement: If BC intersects circle A only at point B, and mCBA = 90°, then CB i

s a​

Mathematics
1 answer:
cricket20 [7]3 years ago
7 0

Line CB touches circle A only at one point (point B), hence CB is a tangent.

<h3>What is a circle?</h3>

A circle is the locus of a point such that its distance from a fixed point (center) is always constant.

When a line intersects a circle in exactly one point the line is said to be tangent to the circle

Hence since Line CB touches circle A only at one point (point B), hence CB is a tangent.

Find out more on circle at: brainly.com/question/24375372

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Nina knows that the average of the x-intercepts represents the line of symmetry for a quadratic function through the x-axis. Whi
Ghella [55]

Answer:

To get the x-intercepts of the function, f(x) = 4x2 – 24x + 20

It has to be equated to zero and the values of x are the x-intercepts. So,

4x2 – 24x + 20 = 0

The resulting equation is a quadratic equation which can be solved by different methods. The solution is

x = 5, 1

The average therefore is:

(1+5/)2 = 3

5 0
3 years ago
Read 2 more answers
Combine like terms for <br> (3x²+2x²-5x+7)+(4x²+2x-3)
horsena [70]

Answer:

9x^2-3x+4

Step-by-step explanation:

(3x^2+2x^2-5x+7)+(4x^2+2x-3)

5x^2-5x+7+4x^2+2x-3

5x^2+4x^2-5x+2x+7-3

9x^2-3x+7-3

9x^2-3x+4

5 0
3 years ago
Read 2 more answers
If the surface area of a cube is 294, whats the length of the cube?
tatyana61 [14]

Answer:

7

Step-by-step explanation:

Total number of surfaces of cube are 6.

Each surface is square.

And area of square is (side)^2

That means, total surface area will be 6 * (side)^2

From question,

294 = 6 * (side)^2 ( Surface area given)

=> 294 ÷ 6 = ( side) ^2

=> 49 = side^2

=> √49 = side

=> 7 = side

So, length of edge of cube is 7

7 0
3 years ago
I need help............​
Alexxandr [17]

Problem 35

The bar graph is shown below

You simply draw various rectangles such that the heights represent the frequency of each animal type.

Eg: there are 20 elephants, so the "elephant" bar is 20 units tall.

You can make the bar graph by hand, or use spreadsheet software. I recommend going with software (if you can).

=======================================================

Problem 36

1 book = 20 mm thickness

5 books = 5*20 = 100 mm thickness

1 paper = 0.016 mm thickness

5 papers = 5*0.016 = 0.08 mm thickness

total thickness = 100+0.08 = 100.08 mm

<h3>Answer: 100.08 mm</h3>

=======================================================

Problem 37

121/11 = 11

121 = 11*11

If we say 11+11+11+...+11, and have 11 copies of these values added, then we get to a sum of 121

This is probably the easiest way to get the answer assuming repeated values are allowed.

You can stop here if your teacher allows you to use repeated values. If not, then move onto the next section.

-----------

If your teacher requires you to add 11 <u>different</u> numbers, then you can follow this procedure

  1. Write out eleven copies of "11" in a sequence
  2. Subtract 2 from the first "11" (to get 9) and add it to the last copy of "11" (to get 13)
  3. Subtract 4 from the second "11" (to get 7) and add it to the second to last copy of "11" (to get 15)
  4. Subtract 6 from the third "11" (to get 5) and add it to the third to last copy of "11" (to get 17)
  5. Subtract 8 from the fourth "11" (to get 3) and add it to the fourth to last copy of "11" (to get 19)
  6. Finally, subtract 10 from the fifth "11" (to get 1) and add it to the fifth to last copy of "11" (to get 21)

After carefully following those steps, you'll get this sequence (in the exact order shown):

{9, 7, 5, 3, 1, 11, 21, 19, 17, 15, 13}

There are three properties to notice of this sequence

  1. It's composed of two decreasing arithmetic sequences {9, 7, 5, 3, 1} and {21, 19, 17, 15, 13}
  2. The 11 in the middle hasn't been changed from the original sequence of nothing but "11"s.
  3. You should find that the terms of this new sequence add to 121.

That sequence we got sorts to {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21}

and we can say 1+3+5+7+9+11+13+15+17+19+21 = 121

<h3>Answer: 1+3+5+7+9+11+13+15+17+19+21 = 121</h3>

5 0
3 years ago
How many four-character passwords can be formed using the characters A, B, C, 1, 2 if the characters can be repeated
olga55 [171]
<h3>Answer: 625</h3>

Work Shown:

The set {A,B,C,1,2} has five items. There are four slots to fill.

So we have 5^4 = 5*5*5*5 = 625 different possible passwords where the characters can be repeated.

7 0
3 years ago
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