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mestny [16]
3 years ago
14

Use this information to complete activities 1.1

Mathematics
1 answer:
Llana [10]3 years ago
5 0
14*4= 56 
so 56 would be the answer 
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If the measurment of the angles of a triangle are 3b,2b,and 4b then what is the smallest angle
d1i1m1o1n [39]

Given:

The measurement of the angles of a triangle are 3b,2b,and 4b.

To find:

The smallest angle.

Solution:

According to the angle sum property, the sum of all angles of a triangle is 180 degrees.

3b+2b+4b=180^{\circ}           [Angle sum property]

9b=180^{\circ}

Divide both sides by 9.

b=20^{\circ}

Now,

3b=3(20^{\circ})=60^{\circ}

2b=2(20^{\circ})=40^{\circ}

3b=4(20^{\circ})=80^{\circ}

Therefore, the smallest angle is 40 degrees.

6 0
3 years ago
What is X-5+10x+3???
Alekssandra [29.7K]

Answer:

-2+11x

Step-by-step explanation:

Combine like terms

(-5+3)+(x+10x)

-5+3 = -2

x+10x = 11x

-2+11x

8 0
3 years ago
Which function represents the relationship between x and y?
zepelin [54]
The correct answer is c
3 0
3 years ago
Can you write for me some formula of log please? I have a pre-test tomorrow about that
Rasek [7]
Y = log(x-5) + log(x+5)
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3 years ago
The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base and PA = 2sqrt5 in. The base edges AB = 10 in, AC
DIA [1.3K]

Answer:

PQ = √84 = 2√21 in ≈ 9.165 in

Step-by-step explanation:

The base edges AB = 10 in, AC = 17 in, and BC = 21 in

Q ∈ BC and AQ is the altitude of the base.

Let BQ = x in ⇒ CQ = (21 - x) in

ΔAQB a right triangle at Q

∴ AQ² = AB² - BQ² = 10² - x²  ⇒(1)

ΔAQC a right triangle at Q

∴ AQ² = AC² - CQ² = 17² - (21-x)²  ⇒(2)

Equating (1) and (2)

∴  10² - x² = 17² - (21-x)²

∴ 10² - x² = 17² - (21² - 42x + x²)

∴  10² - x² = 17² - 21² + 42x - x²

∴ 10² - 17² + 21² = 42x

∴ 42x = 252

∴ x = 252/42 = 6

Substitute at (1)

∴ AQ² = AB² - BQ² = 10² - x² = 100 - 36 = 64 = 8²

∴ AQ = 8 in

∵ The lateral edge PA of a triangular pyramid ABCP is perpendicular to the base

∴ PA⊥AQ

∴ ΔPAQ is a right triangle at A,

PA = 2sqrt5 in  and AQ = 8 in

∴ PQ (hypotenuse) = √(PA² + AQ²)

∴ PQ = √[(2√5)² + 8²] = √(20+64) = √84 = 2√21 in ≈ 9.165 in

6 0
3 years ago
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