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Wittaler [7]
3 years ago
10

Given: PQ= TQ, UQ: QS

Mathematics
1 answer:
joja [24]3 years ago
3 0

Answer:

The required prove is shown below.

Step-by-step explanation:

Consider the provided statement.

The required figure is shown below:

 Statement                                             Reasons

PQ ≅ TQ; UQ ≅ QS                               Given

PQ = TQ, UQ = QS                     Definition of Congruence

PQ + QS = PS; TQ+QU = TU    Segment Addition Postulate

TQ + QS = PS                            Substitution  (PQ = TQ)

TQ + QS = TU                            Substitution  (QU = QS)

PS = TU                                    Transitive Property

Prove: PS ≅TU                          Definition of Congruence

Hence, the required prove is shown above.

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Answer:

1/3 I believe because we already know 2 of them are hearts

Step-by-step explanation:

3 0
3 years ago
The perimeter of a rectangular field is 60 meters. Its area is 200 square meters. Find its dimensions.
Vanyuwa [196]

Answer:

20 m by 10 m

Step-by-step explanation:

let w be width and l be length , then

2(l + w) = 60 ( divide both sides by 2 )

l + w = 30 ( subtract w from both sides )

l = 30 - w → (1)

lw = 200 → (2)

Substitute l = 30 - w into (2)

w(30 - w) = 200 ← distribute parenthesis on left side

30w - w² = 200 ( subtract 200 from both sides )

30w - w² - 200 = 0 ( multiply through by - 1 )

w² - 30w + 200 = 0 ← in standard form

(w - 10)(w - 20) = 0 ← in factored form

Equate each factor to zero and solve for w

w - 10 = 0 ⇒ w = 10

w - 20 = 0 ⇒ w = 20

Substitute these values into (1)

l = 30 - 10 = 20

l = 30 - 20 = 10

dimensions of field is 20 m by 10 m

4 0
2 years ago
Read 2 more answers
Which of the following are identities? Check all that apply
Natasha2012 [34]

Answer:

A, C

Step-by-step explanation:

Actually, those questions require us to develop those equations to derive into trigonometrical equations so that we can unveil them or not. Doing it only two alternatives, the other ones will not result in Trigonometrical Identities.

Examining

A) True

\frac{1-tan^{2}x}{2tanx} =\frac{1}{tan2x} \\ \frac{1-tan^{2}x}{2tanx} =\frac{1}{\frac{2tanx}{1-tan^{2}x}}\\ tan2x=\frac{1-tan^{2}x}{2tanx}

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B) False,

No further development towards a Trig Identity

C) True

Double Angle Sine Formula sin2\alpha =2sin\alpha *cos\alpha

sin(8x)=2sin(4x)cos(4x)\\2sin(4x)cos(4x)=2sin(4x)cos(4x)

D) False No further development towards a Trig Identity

[sin(x)-cos(x)]^{2} =1+sin(2x)\\ sin^{2} (x)-2sin(x)cos(x)+cos^{2}x=1+2sinxcosx\\ \\sin^{2} (x)+cos^{2}x=1+4sin(x)cos(x)

7 0
3 years ago
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4 - 2(x-4)/ 3= 3(2x+5)/2<br>pls answer fast​
omeli [17]

Answer:

- (5/22)

Step-by-step explanation:

(12-2x+8)/3 = (6x + 15)/2

(20 - 2x)/3 = ( 6x + 15)/2

Now cross multiplication

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3 0
2 years ago
What is the radius for the circle given by the equation (x - 2)^2 + y^2 = 14?
Anarel [89]

Easily, take the square root of the number in the right side to find the radius

\sqrt{14}

is the answer

7 0
2 years ago
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