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3 years ago

Which of the following is the correct name for LM in ABC below

Mathematics

2 answers:

Lena [83]3 years ago

6 0

FM is midsegment of triangle ABC

Answer

C. Midsegment

Irina-Kira [14]3 years ago

5 0

Answer:

C) midsegment

Step-by-step explanation:

An angle bisector cuts an angle in half. LM does not do this.

An altitude extends from a vertex perpendicular to the opposite side. LM does not do this.

A midsegment goes from the midpoint of one side to the midpoint of another side. LM does this.

A perpendicular bisector splits a side in half, forming a right angle. LM does not do this.

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Let f(x) = 3x + 2 and g(x) =x - 3. Find f(x) - g(x).

Arturiano [62]

2x + 5 is your answer

8 0

3 years ago

Read 2 more answers

3 100° B. 48 270 o ZA= degrees b= (=

solong [7]

Answer:

Angle A = 53

Step-by-step explanation:

All triangles add up to 180 so...

Lets use x to describe angle A

100+27+x=180

127+x=180

x=53

I'm confused what its asking about b and c but hope this helps a bit :)

6 0

3 years ago

Still I need sleep...

Pani-rosa [81]

Answer:

d) 28.472

Step-by-step explanation:

you need to find the length of all the outside edges. for the triangles you will need to use the pythagorean theorem

4 0

3 years ago

3 regions are defined in the figure find the volume generated by rotating the given region about the specific line

anastassius [24]

The volume generated by rotating the given region $R_{3}$ about OC is $\frac{4}{g} \pi$

<h3>Washer method</h3>

Because the given region ( $R_{3}$ ) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.

solution

We first find the value of x and y

$y=2(x)^{\frac{1}{4} }$

$x=(\frac{y}{2} )^{4}$

$y=2x$

$x=\frac{y}{2}$

$\int\limits^a_b {\pi } \, (R_{o^{2} } - R_{i^{2} } ) dy$

$R_{o} = x = \frac{y}{2}$

$R_{i} = x= (\frac{y}{2}) ^{4}$

$a=0, b=2$

$v= \int\limits^2_o {\pi } \, [(\frac{y}{2})^{2} - ((\frac{y}{2}) ^{4} )^{2} ) dy$

$v= \pi \int\limits^2_o= [\frac{y^{2} }{4} - \frac{y^{8} }{2^{8} }} ] dy$

$v= \pi [\int\limits^2_o {\frac{y^{2} }{4} } \, dy - \int\limits^2_o {\frac{y}{2^{8} } ^{8} } \, dy ]$

$v=\pi [\frac{1}{4} \frac{y^{3} }{3} \int\limits^2_0 - \frac{1}{2^{8} } \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi$

A similar question about finding the volume generated by a given region is answered here: brainly.com/question/3455095

6 0

2 years ago

What is the total amount paid on a 30 year mortgage for $53,000 with an interest rate of 4.25%

Sonja [21]

Answer:

67,575.00

Step-by-step explanation:

4 0

3 years ago