All categories

3 years ago

Which of the following pair(s) of circles have l as a common internal tangent? Select ALL that apply.

Mathematics

2 answers:

koban [17]3 years ago

8 0

Answer:

A and B

A and C

Step-by-step explanation:

Common internal tangents intersect the segment joining the centers of the two circles.

Tangent I is common internal tangent to circles A and B as the line joining the centers of circles will intersect the tangent I.

Tangent I is also a common internal tangent to circle A and C.The line joining the centers of circles A and C intersects tangent I .

Tangent I is not a common internal tangent to circles B and C,

kkurt [141]3 years ago

7 0

It looks like only A and B, but could be trick question?

You might be interested in

Marlene noticed that when a pair of parallel lines was cut by a transversal, two particular vertical angles were reflections of

Trava [24]

A.the measures of the angles must be equal
Vertical Angles- are the angles opposite of each other when two lines cross. They are always equal.

3 0

2 years ago

Read 2 more answers

Evaluate. −8/125−−−−√3<br> A. -4/25<br> B. −2/5<br> C. 4/25<br> D. 2/5

larisa [96]

The correct question is: evaluate

$\sqrt[3]{- \frac{8}{125} }$

The answer is: option B. -2 / 5

Explanation:

$\sqrt[3]{- \frac{8}{125} }=- \frac{ \sqrt[3]{8} }{ \sqrt[3]{125} } =- \frac{ \sqrt[3]{2^3} }{ \sqrt[3]{5^3} } =- \frac{2}{5}$

4 0

3 years ago

Find the vector that has the same direction as 3, 2, −6 but has length 2.

Nata [24]

Answer:

The vector is $\vec r = \left(\frac{6}{7},\frac{4}{7},-\frac{12}{7}\right)$ .

Step-by-step explanation:

We can determine the equivalent vector ( $\vec r$ ), dimensionless, by means of the following formula:

$\vec r = \frac{\vec u}{\|\vec u\|} \cdot \|\vec r\|$ (1)

Where:

$\vec u$ - Original vector, dimensionless.

$\|\vec u\|$ - Norm of the original vector, dimensionless.

$\|\vec r\|$ - Norm of the new vector, dimensionless.

The norm of the original vector is determined by the following definition:

$\|\vec u\| = \sqrt{\vec u\,\bullet \,\vec u}$ (2)

If we know that $\vec u = (3, 2, -6)$ , then the norm of the original vector is:

$\|\vec u\| = \sqrt{(3)\cdot (3)+(2)\cdot (2)+(-6)\cdot (-6)}$

$\|\vec u\| = 7$

If we know that $\|\vec r\| = 2$ , then the new vector is:

$\vec r = \frac{2}{7}\cdot (3,2,-6)$

$\vec r = \left(\frac{6}{7},\frac{4}{7},-\frac{12}{7}\right)$

The vector is $\vec r = \left(\frac{6}{7},\frac{4}{7},-\frac{12}{7}\right)$ .

7 0

3 years ago

I need help with this graph?!

Anastasy [175]

Answer:

Im pretty sure its D. all numbers greater than or equal to -3

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

26 POINTS WILL GIVE BRAINLIEST

Zina [86]

2: They do not because the hypotenuse is larger and a^2 + b^2
3. The tiangle is acute because it has all angles smaller than 90*
12: The buillding is approximately 100 ft tall give or take. I am sorry If i am wrong on the last one

8 0

3 years ago