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Leno4ka [110]
3 years ago
12

This triangle has one side that lies on an extended line segment.

Mathematics
1 answer:
monitta3 years ago
7 0

Answer:

65°

Step-by-step explanation:

x = 180 - 44 - (180-109) = 109 - 44 = 65°

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(06.01)<br> Solve x + 3y = 9<br> 3x − 3y = −13
Irina-Kira [14]

Answer:

Slove For X

x= -1

Solve for Y

Y= 10/3

Step-by-step explanation:

SO I did Elimination method

To isolate X and to solve for it

x+3y=9

3x−3y=−13

Add these equations to eliminate y:

4x=−4

Solve for X

x=-1

Now plug x back in to solve for y

4 0
3 years ago
Question is in the picture (surface area)
Marina CMI [18]

Answer:

Your answer is 62 square inches

Step-by-step explanation:

You can cross off 1 & 2, as they are too small.

Lets split this up.

(2*3)+(2*3)+(3*5)+(3*5)+(2*5)+(2*5)

2*3 + 2*3 = 12

3*5 + 3*5 = 30

2*5 + 2*5 = 20

30+20+12 = 62

7 0
2 years ago
HELP Use either law of sines or law of cosine. Need help on this problem! show work please!​
Marina86 [1]

Answer: x = 15.035677095729 approximately

Round this however you need to.

=================================================

Explanation:

I'm assuming you want to find the value of x, which your diagram is showing to be the length of segment QR.

If so, then we'll need to find the measure of angle Q first. Using the law of sines, we get the following:

sin(Q)/q = sin(R)/r

sin(Q)/PR = sin(R)/PQ

sin(Q)/13 = sin(85)/19

sin(Q) = 13*sin(85)/19

sin(Q) = 0.6816068987

Q = arcsin(0.6816068987) ... or ... Q = 180-arcsin(0.6816068987)

Q = 42.9693397461 ... or ... Q = 137.0306602539

These values are approximate.

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

Now if Q = 42.9693397461 approximately, then angle P is

P = 180-Q-R

P = 180-42.9693397461-85

P = 52.0306602539

Similarly, if Q = 137.0306602539 approximately, then,

P = 180-Q-R

P = 180-137.0306602539-85

P = -42.0306602539

A negative angle is not possible, so we'll ignore Q = 137.0306602539

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

The only possible value of angle P is approximately P = 52.0306602539

Let's apply the law of sines again to find side p, aka segment QR

sin(P)/p = sin(R)/r

sin(P)/QR = sin(R)/PQ

sin(52.0306602539)/x = sin(85)/19

19*sin(52.0306602539) = x*sin(85)

19*sin(52.0306602539)/sin(85) = x

x = 15.035677095729

This value is approximate.

Round this value however you need to.

4 0
3 years ago
Read 2 more answers
Integral of cosecx dx =log|tanx/2| show that.​
Pie

ANSWER:

Let t = logtan[x/2]

⇒dt = 1/ tan[x/2] * sec² x/2 × ½ dx

⇒dt = 1/2 cos² x/2 × cot x/2dx

⇒dt = 1/2 * 1/ cos² x/2 × cosx/2 / sin x/2 dx

⇒dt = 1/2 cosx/2 / sin x/2 dx

⇒dt = 1/sinxdx

⇒dt = cosecxdx

Putting it in the integration we get,

∫cosecx / log tan(x/2)dx

= ∫dt/t

= log∣t∣+c

= log∣logtan x/2∣+c where t = logtan x/2

7 0
2 years ago
Help with this question ??
salantis [7]

Answer:

15 units per unit

Step-by-step explanation:

From the graph we can easily tell that y increases from 0 to 15 as x increases from 2 to 3.  

Thus, the average rate of change of f(x) on the interval [2, 3] is:

    1         15 - 0       15 units

--------- * ----------- = -------------- = 15 units (rise) per unit (run).

 3 - 2          1                 1

6 0
3 years ago
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