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Colt1911 [192]
3 years ago
9

I need some help, with my algebra ​

Mathematics
2 answers:
quester [9]3 years ago
7 0

Answer:

P=24

A=24

Step-by-step explanation:

To get the perimeter and area of this figure, first we would have to get the value of c using the Pythagorean theorem.

a^2+b^2=c^2

(6)^2+(8)^2=c^2

36+64=c^2

100=c^2

10=c

Perimeter= a+b+c

P=6+8+10

P=14+10

P=24

A=1/2*b*h

A=1/2*6*8

A=3*8

A=24

Hope this helps ;) ❤❤❤

Viktor [21]3 years ago
5 0

Answer:

Hey there!

Perimeter: 24

Area: 24

36+64=x^2

x=10

Perimeter: 10+6+8=24

Area: 1/2bh, 1/2(48)=24

Both the area and perimeter are 24.

Hope this helps :)

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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.

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