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lubasha [3.4K]
4 years ago
11

If quadrilateral PQRS is a kite, which statements must be true? Check all that apply.

Mathematics
2 answers:
kow [346]4 years ago
7 0

Answer: The true statements are,

QP ≅ QR

PM ≅ MR

∠QPS ≅ ∠QRS

Step-by-step explanation:

A kite, is a quadrilateral which has two pairs of congruent sides.

Also, In kite the major diagonal bisects the minor diagonal perpendicularly.

Here, PQRS is a kite, in which,

Sides PQ and PS are congruent  to sides QR and RS respectively,

That is, QP ≅ QR and PS ≅ RS

Since, QS ≅ QS ( Reflexive )

By SSS postulate of congruence,

Δ PQS ≅ Δ RQS

By CPCTC,

∠ QPR ≅ ∠ QPS

Also, QS is the major diagonal and PR is the minor diagonal,

⇒ QS bisects PR

⇒ PM ≅ MR ( Where M is the intersection point of these diagonals. )

⇒ First second and fifth options are correct.

mel-nik [20]4 years ago
3 0

A kite is a quadrilateral in which two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means that one side can’t be used in both pairs).

Then:

  • Two disjoint pairs of consecutive sides are congruent by definition - QP≅QR and QR is not congruent to RS (one side can’t be used in both pairs);
  • One diagonal (segment QS, the main diagonal) is the perpendicular bisector of the other diagonal (segment PR, the cross diagonal), so PM≅MR;
  • The opposite angles at the endpoints of the cross diagonal are congruent, thus ∠QPS≅∠QRS.
  • ∠PQR is not congruent to ∠PSR, because they are not angles at the endpoints of the cross diagonal.

Answer: correct options are A, B and E.

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Step-by-step explanation:

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\frac{\sqrt[4]{3}}{\sqrt[4]{2x}}=\frac{\sqrt[4]{3(2^3x^3)}}{\sqrt[4]{2x(2^3x^3)}}

Simplifiying, you get:

\frac{\sqrt[4]{3(8x^3)}}{\sqrt[4]{2x(2^3x^3)}}=\frac{\sqrt[4]{24x^3}}{\sqrt[4]{2^4x^4}}=\frac{\sqrt[4]{24x^3}}{2x}

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4 years ago
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---

slope-intercept form:

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Read 2 more answers
The half life of uranium-232 is 68.9 years. (show step by step)
dalvyx [7]

Answer:

1.) 8.09g ; 2) 206.7 years

Step-by-step explanation:

Given the following :

Half-life(t1/2) of Uranium-232 = 68.9 years

a) If you have a 100 gram sample, how much would be left after 250 years?

Initial quantity (No) = 100g

Time elapsed (t) = 250 years

Find the quantity of substance remaining (N(t))

Recall :

N(t) = No(0.5)^(t/t1/2)

N(250) = 100(0.5)^(250/68.9)

N(250) = 100(0.5)^3.6284470

N(250) = 100 × 0.0808590

= 8.0859045

= 8.09g

2) If you have a 100 gram sample, how long would it take for there to be 12.5 grams remaining?

Using the relation :

N / No = (1/2)^n

Where N = Amount of remaining or left

No = Original quantity

n = number of half-lifes

N = 12.5g ; No = 100g

12.5 / 100 = (1/2)^n

0.125 = (1/2)^n

Converting 0.125 to fraction

(1/8) = 1/2^n

8 = 2^n

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n = 3

Recall ;

Number of half life's (n) = t / t1/2

t = time elapsed ; t1/2 = half life

3 = t / 68.9

t = 3 × 68.9

t = 206.7 years

7 0
3 years ago
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