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Mashutka [201]
3 years ago
15

HELP PLEASEEEEEEEEEEEEEEEEEE

Mathematics
2 answers:
harina [27]3 years ago
3 0
The answer you seek is 5. Enjoy 
Serga [27]3 years ago
3 0

Answer: The value of 'n' is 6.

Step-by-step explanation:  Given that ΔXWZ is an isosceles triangle, where XZ = n, WZ = 2n - 5, and the perimeter of the triangle is 20 units.

We are to find the value of 'n'.

As given in the figure, in ΔXWZ,

WX = WZ.

So, WX = 2n - 5.

We know that the perimeter of a triangle is equal to the sum of its three sides, so the perimeter of ΔXWZ is given by

P=XZ+WX+WZ=n+(2n-5)+(2n-5)=5n-10.

According to the question, we have

P=20\\\\\Rightarrow 5n-10=20\\\\\Rightarrow 5n=30\\\\\Rightarrow n = 6.

Thus, the value of 'n' is 6.

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Determine the values of the unlabeled parts of the triangle.
____ [38]

Answer:

∠Q = 63.4º, PQ = 2, QO = 4.47

Step-by-step explanation:

A triangle is a polygon with three sides and three angles. There are different types of triangles such as the equilateral triangle, isosceles triangle, scalene triangle, right angled triangle.

Triangle POQ is a right angled triangle because ∠P = 90°. Also, ∠O = 26.6°.

Therefore in ΔPOQ: ∠P + ∠O + ∠Q = 180° (sum of angles in a triangle).

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We can find PQ and QO using sine rule. Sine rule for ΔPOQ is given as:

\frac{PO}{sin Q}= \frac{PQ}{sinO}= \frac{QO}{sin P}\\\\Finding\ PQ  :\\\\\frac{PO}{sin Q}= \frac{PQ}{sinO}\\\\\frac{4}{sin(63.4)} =\frac{PQ}{sin(26.6)}\\\\PQ= \frac{4}{sin(63.4)} *sin(26.6)\\\\PQ = 2\\\\Also\ finding\ QO:\\\\\frac{PO}{sin Q}= \frac{QO}{sinP}\\\\\frac{4}{sin(63.4)} =\frac{QO}{sin(90)}\\\\QO=\frac{4}{sin(63.4)} *sin(90)\\\\QO = 4.47

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4 years ago
HELP NEEDED PLEASEEE:(
poizon [28]

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four more

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is

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