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eduard
2 years ago
12

By which rule are these triangles congruent?

Mathematics
1 answer:
Hunter-Best [27]2 years ago
5 0

Answer:

A. AAS

Step-by-step explanation:

In both triangles, the line mn and jh are congruent, the angles k and g are congruent, and h and m are congruent.

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What's the area of the composite figure below ?
JulsSmile [24]

Answer:

31.13

Step-by-step explanation:

If you look carefully, you will find a semicircle and a triangle.

The total area is =

\frac{\pi r^{2}}{2} + \frac{1}{2} \times base \times height\\

  • r = 4 - 0 = 4
  • base = |-4-0 | = 4
  • height = 14 - 9 = 5

So if we put the values:

\frac{\pi \times 4^{2}}{2} + \frac{1}{2} \times 4 \times 3\\= 31.13

3 0
3 years ago
1-sin^2x/sinx-cscx <br><br>How do I solve this problem? I keep getting the wrong answer.
Annette [7]

Answer:

-sinx

Step-by-step explanation:

a trig identity that is crucial to solving this problem is: sin^2 + cos^2 = 1

with knowing that, you can manipulate that and turn it into 1 - sin^2x = cos^x

so 1-sin^2x/sinx - cscx becomes cos^2x/sinx - cscx

it is also important to know that cscx is the same thing as 1/sinx

knowing this information, cscx can be replaced with 1/sinx

(cos^2x)/(sinx - 1/sinx)

now sinx and 1/sinx do not have the same denominator, so we need to multiply top and bottom of sinx by sinx; it becomes....

cos^2x

---------------------

(sin^2x - 1)/sinx

notice how in the denominator it has sin^2x-1 which is equal to -cos^2x

so now it becomes:

cos^2x

--------------

-cos^2x/sinx

because we have a fraction over a fraction, we need to flip it

cos^2x          sinx

---------- * ----------------

1                  -  cos^2x

because the cos^2x can cancel out, it becomes 1

now the answer is -sinx

3 0
3 years ago
Please help!!!!!!! How many inches would a point on the outer edge of the large gear travel in a 150º rotation? (The gear has a
damaskus [11]
In a full rotation, 360°, a point on the edge of a gear would travel a distance equal to its circumference.  The circumference of a circle is:

C=2πr, and since we are only traveling 150°, we need to set up an appropriate ratio for circumference to the distance the point travels:

d/C=150/360

d=5C/12 and since C=2πr

d=10πr/12

d=5πr/12, and since r=4in

d=5*4π/12 in

d=20π/12 in

d=5π/3 in

d≈5.24 in (to nearest hundredth of an inch)
4 0
3 years ago
Use the diagram that awnser the questions.
CaHeK987 [17]
Part A
 The first thing we must do in this case is to hide the slopes of each line.
 line m:
 m = (- 4-3) / (0 - (- 4))
 m = -7 / 4
 Line n:
 n = (- 2-2) / (3-1)
 n = -4 / 2
 n = -2
 Answer:
 Lines m and n are not parallel because their slopes are different.

 Part B:
 
We look for the slope of the K line:
 k = (1 - (- 3)) / (4 - (- 3))
 k = 4/7
 We observe that it is true that:
 k = -1 / m
 Answer: 
 The lines are perpendicular.
8 0
3 years ago
Read 2 more answers
Find the arc length of the following problem:​
Harrizon [31]

Answer:

61.261 {ft}^{2}

Step-by-step explanation:

\frac{270}{360}  \times 2\pi \: r \\  \frac{270}{360}  \times 2 \times \pi \times 13 \\  = 61.261 {ft}^{2}

<h3>Can I have the brainliest please?</h3>
5 0
3 years ago
Read 2 more answers
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