What is the area of the triangle?

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Svet_ta [14]

I’m sorry i don’t speak Spanish!

7 0

3 years ago

If jill climed 100 feet more how many feet above sea level would she be

Anit [1.1K]

You have to have a better question to answer that, 100 feet more what?

5 0

3 years ago

Please help I’m really confused and need to learn this

egoroff_w [7]

Answer:

2.

Step-by-step explanation:

For #2, another way to word this question is: For which of the following trig functions is π/2 a solution? Well, go through them one by one. If you plug π/2 into sinθ, you get 1. This means that when x is π/2, y is 1. Try and visualize that. When y is 1, that means you moved off the x-axis; so y = sinθ is NOT one of those functions that cross the x-axis at θ = π/2. Go through the rest of them. For y = cos(π/2), you get 0. At θ = π/2, this function crosses the x-axis. For y = tanθ, your result is undefined, so that doesn't work. Keep going through them. You should see that y = secθ is undefined, y = cscθ returns 1, and y = cotθ returns 0. If you have a calculator that can handle trig functions, just plug π/2 into every one of them and check off the ones that give you zero. Graphically, if the y-value is 0, the function is touching/crossing the x-axis.

Think about what y = secθ really means. It's actually y = 1/(cosθ), right? So what makes a fraction undefined? A fraction is undefined when the denominator is 0 because in mathematics, you can't divide by zero. Calculators give you an error. So the real question here is, when is cosθ = 0? Again, you can use a calculator here, but a unit circle would be more helpful. cosθ = π/2, like we just saw in the previous problem, and it's zero again 180 degrees later at 3π/2. Now read the answer choices.

All multiples of pi? Well, our answer looked like π/2, so you can skip the first two choices and move to the last two. All multiples of π/2? Imagine there's a constant next to π, say Cπ/2 where C is any number. If we put an even number there, 2 will cut that number in half. Imagine C = 4. Then Cπ/2 = 2π. Our two answers were π/2 and 3π/2, so an even multiple won't work for us; we need the odd multiples only. In our answers, π/2 and 3π/2, C = 1 and C = 3. Those are both odd numbers, and that's how you know you only need the "odd multiples of π/2" for question 3.

6 0

3 years ago

CAN SOMEONE PLEASE PLEASE PLEASE HELP ME, YOU’LL GET FREE EASY POINTS IF YOU GIVE ME THE RIGHT ANSWER !!

bekas [8.4K]

Answer:

reflection across BC
the image of a vertex will coincide with its corresponding vertex
SSS: AB≅GB, AC≅GC, BC≅BC.

Step-by-step explanation:

We want to identify a rigid transformation that maps congruent triangles to one-another, to explain the coincidence of corresponding parts, and to identify the theorems that show congruence.

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Triangles GBC and ABC share side BC. Whatever rigid transformation we use will leave segment BC invariant. Translation and rotation do not do that. The only possible transformation that will leave BC invariant is <em>reflection across line BC</em>.

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In part 3, we show ∆GBC ≅ ∆ABC. That means vertices A and G are corresponding vertices. When we map the congruent figures onto each other, <em>corresponding parts are coincident</em>. That is, vertex G' (the image of vertex G) will coincide with vertex A.

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The markings on the figure show the corresponding parts to be ...

side AB and side GB
side AC and side GC
angle ABC and angle GBC
angle BAC and angle BGC

And the reflexive property of congruence tells us BC corresponds to itself:

side BC and side BC

There are four available congruence theorems applicable to triangles that are not right triangles

SSS -- three pairs of corresponding sides
SAS -- two corresponding sides and the angle between
ASA -- two corresponding angles and the side between
AAS -- two corresponding angles and the side not between

We don't know which of these are in your notes, but we do know that all of them can be used. AAS can be used with two different sides. SAS can be used with two different angles.

SSS

Corresponding sides are listed above. Here, we list them again:

AB and GB; AC and GC; BC and BC

SAS

One use is with AB, BC, and angle ABC corresponding to GB, BC, and angle GBC.

Another use is with BA, AC, and angle BAC corresponding to BG, GC, and angle BGC.

ASA

Angles CAB and CBA, side AB corresponding to angles CGB and CBG, side GB.

AAS

One use is with angles CBA and CAB, side CB corresponding to angles CBG and CGB, side CB.

Another use is with angles CBA and CAB, side CA corresponding to angles CBG and CGB, side CG.

3 0

1 year ago

Read 2 more answers

In ΔABC, point M is the midpoint of side AB and point D is the midpoint of segment MC. Prove that the area of ΔADC= the area of

Anna11 [10]

A line segment from a vertex to the midpoint of the opposite side is a "median". A median divides the area of the triangle in half, as it divides the base in half without changing the altitude.
AAMC is half AABC. AADC is half AAMC, so is 1/4 of AABC. (By the formula for area of a triangle.)
ABMC is half AABC. ABMD is half ABMC, so is 1/4 of AABC. (By the formula for area of a triangle.)
Then, AADC = 1/4 AABC = ABMC, so AADC = ABMC by the transitive property of equality.

8 0

3 years ago