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Dominik [7]
3 years ago
11

Consider in the figure below.

Mathematics
2 answers:
Dahasolnce [82]3 years ago
7 0

Answer:

UV = 148

VD = 78

TC = 72

Step-by-step explanation:

BD is the perpendicular bisector of side UV.

Therefore, ΔUDV is an isosceles triangle.

This implies that UD = VD and BV = UB so UV = 2 x BV

  • Given that UD = 78, and UD = VD, then VD = 78
  • Given that BV = 74, and BV = UB, then UV = 2 x 74 = 148

ΔUDC is a right triangle.

Given CD = 30 and UD = 78,

and using Pythagoras' Theorem, we can calculate UC:

UC = √(UD² - CD²)

⇒ UC = √(78² - 30²)

⇒ UC = 72

CD is the perpendicular bisector of side UT.

Therefore, ΔUDT is an isosceles triangle, so UC = TC

Since UC = 72, then TC = 72

yawa3891 [41]3 years ago
6 0

Here BD is perpendicular bisector

So

  • UB=BV=74

\\ \tt\hookrightarrow UV=74+74=148

Apply Pythagorean theorem

\\ \tt\hookrightarrow BD^2=UD^2-UB^2=UD^2-VD^2

  • UD=VD=78=TD

\\ \tt\hookrightarrow TC^2=TD^2-CD^2

\\ \tt\hookrightarrow TC^2=78^2-30^2

\\ \tt\hookrightarrow TC^2=6084-900=5184

\\ \tt\hookrightarrow TC=72

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What is the prime factorization of 32?
Dahasolnce [82]

Answer:

  2^5

Step-by-step explanation:

Keep dividing by primes until you can't anymore. The numbers you divided by are the prime factors. The only one you need is 2.

  32 = 2·16 = 2·2·8 = 2·2·2·4 = 2·2·2·2·2

  32 = 2^5

7 0
3 years ago
The sum of three consecutive odd integers is − 87 . find the numbers.
DIA [1.3K]
Three consecutive odd integers are n, n+2 and n+4

n + n+2 + n+4 = -87
3n + 6 = -87
3n = -87 - 6
3n = -93
n = -93/3
n = -31

n+2 = -31 + 2 = -29
n+4 = -31 + 4 = -27

The numbers are -31, -29 and -27
5 0
4 years ago
Heya!
nataly862011 [7]

Answer:

m\angle QSN=65^\circ

Step-by-step explanation:

In the given figure, PQRS is a rhombus and SRM is an equilateral triangle.

We are also given that SN⊥RM and that ∠PRS = 55°.

And we want to find the measure of ∠QSN.

Remember that since PQRS is a rhombus, the angles formed by its diagonals are right angles. Let the intersection point of the diagonals be K. Therefore:

m\angle RKS=90^\circ

Now, RKS is also a triangle. The interior angles of all triangles must be 180. Thus:

m\angle RKS+m\angle KSR+m\angle SRK=180

Substitute in known values:

90+55+m\angle KSR=180

Solve for ∠KSR:

m\angle KSR+145=180\Rightarrow m\angle KSR=35^\circ

Since SRM is an equilateral triangle, this means that:

m\angle SRM=m\angle RMS=m\angle MSR=60^\circ

Note that RNS is also a triangle. Therefore:

m\angle SRM+m\angle RNS+m\angle NSR=180

Substitute in known values:

60+90+m\angle NSR=180

So:

m\angle NSR+150=180\Rightarrow m\angle NSR=30^\circ

∠QSN is the addition of the two angles:

m\angle QSN=m\angle KSR+m\angle NSR

Therefore:

m\angle QSN=35+30=65^\circ

4 0
3 years ago
Read 2 more answers
Can someone give me an explanation on how to do it
VLD [36.1K]

Its C for the answer all it wants is an overlapping congruent

6 0
4 years ago
Please help me with this problem!
aksik [14]
3/4 equals 15/20, so b+3=15. B=12
6 0
3 years ago
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