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3 years ago

In circle M, segment AB is tangent to the circle at point C. AB has endpoints such that AM BM  ,

Mathematics

1 answer:

iren2701 [21]3 years ago

5 0

Answer:

Step-by-step explanation:

We can use the geometric mean theorem:

The altitude on the hypotenuse is the geometric mean of the two segments it creates.

In your triangle, the altitude is the radius CM and the segments are AC and BC.

$CM = \sqrt{AC \times BC} = \sqrt{ 9 \times 4} = \sqrt{36} = \mathbf{6}\\\text{The radius of the circle M is $\large \boxed{\mathbf{6}}$}$

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When 3x2-9x + 3 is factored, the result is 3(x2-3x + 1). Explain why it is necessary to include the term 1 in this factored form

solong [7]

It is necessary to include the term one then you will have the plus three

7 0

3 years ago

4a^5-2a^5+4b+b How to combine this into likewise terms (^number stands for exponents) need help today pls!!!

Nat2105 [25]

Answer: 2a^5+5b

Step-by-step explanation:

This is actually pretty simple

The exponent with the same power can be subtracted so

4a^5-2a^5 is 2a^5 like 4x-2x which is equal to 2x

4b+b is equal to 5b

2a^5+5b is your answer

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3 years ago

Daniel puts some wheat flour into an empty bowl. Then he add 2/3 cup rye flour to make a total of 2 5/12 cups of flour. How much

lozanna [386]

Answer:

1 3/4 cups of wheat flour is in the bowl

Step-by-step explanation:

x + 2/3 = 2 5/12

First, convert the 2/3 to 12ths by multiplying numerator and denominator by 4 = 8/12

Then convert the 2 5/12 to an improper fraction 29/12

x + 8/12 = 29/12

subtract 8/12 from both sides

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6 0

2 years ago

What does the fundamental theorem of algebra illustrate?

UNO [17]

Answer:

The fundamental theorem of algebra guarantees that a polynomial equation has the same number of complex roots as its degree.

Step-by-step explanation:

The fundamental theorem of algebra guarantees that a polynomial equation has the same number of complex roots as its degree.

We have to find the roots of this given equation.

If a quadratic equation is of the form $ax^{2}+bx +c=0$

Its roots are $\frac{-b+\sqrt{b^{2}-4ac } }{2a}$ and $\frac{-b-\sqrt{b^{2}-4ac } }{2a}$

Here the given equation is $2x^{2}-4x-1$ = 0

a = 2

b = -4

c = -1

If the roots are $x_{1} and x_{2}$ , then

$x_{1}$ = $\frac{-2+\sqrt{(-4)^{2}-4\times 2\times (-1)}}{2\times 2}$

= $\frac{4 +\sqrt{24}}{4}$

= $\frac{2+\sqrt{6} }{2}$

$x_{2}$ = $\frac{-2-\sqrt{(-4)^{2}-4\times 2\times (-1)}}{2\times 2}$

= $\frac{4 +\sqrt{8}}{4}$

= $\frac{2-\sqrt{6} }{2}$

These are the two roots of the equation.

6 0

3 years ago

What is are SLOPE and EQUATION of the image?

zepelin [54]

Answer:

y=3x

Step-by-step explanation:

y=mx+b

y=3x+0

y=3x

8 0

3 years ago