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dalvyx [7]
3 years ago
9

Quadrilateral P Q R S is shown. Diagonals are drawn from point Q to point S and from point P to point R and intersect at point M

. Sides P S and R S are congruent. If quadrilateral PQRS is a kite, which statements must be true? Select three options QP ≅ QR PM ≅ MR QR ≅ RS ∠PQR ≅ ∠PSR ∠QPS ≅ ∠QRS
Mathematics
2 answers:
kobusy [5.1K]3 years ago
7 0
Your anwser for your question is QP Qr and Qrs
AysviL [449]3 years ago
4 0

Answer:

QP = QR

PM = MR

QRS = QPS

Step-by-step explanation:

took the test haha

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Margaret [11]

Answer:

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3 years ago
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Mandarinka [93]

Step-by-step explanation:

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6 0
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Read 2 more answers
In a G.P the 3rd term is 4 times the 1st term and the sum of the 2nd term and the 4th term is 30.find the common ratio,find the
STALIN [3.7K]

Answer:

the common ratio is either 2 or -2.

the sum of the first 7 terms is then either 765 or 255

Step-by-step explanation:

a geometric sequence or series of progression (these are the most common names for the same thing) means that every new term of the sequence is created by multiplying the previous term by a constant factor which is called the common ratio.

so,

a1

a2 = a1×f

a3 = a2×f = a1×f²

a4 = a3×f = a1×f³

the problem description here tells us

a3 = 4×a1

and from above we know a3 = a1×f².

so, f² = 4

and therefore the common ratio = f = 2 or -2 (we need to keep that in mind).

again, the problem description tells us

a2 + a4 = 30

a1×f + a1×f³ = 30

for f = 2

a1×2 + a1×2³ = 30

2a1 + 8a1 = 30

10a1 = 30

a1 = 3

for f = -2

a1×-2 + a1×(-2)³ = 30

-10a1 = 30

a1 = -3

the sum of the first n terms of a geometric sequence is

sn = a1×(1 - f^(n+1))/(1-f) for f <>1

so, for f = 2

s7 = 3×(1 - 2⁸)/(1-2) = 3×-255/-1 = 3×255 = 765

for f = -2

s7 = -3×(1 - (-2)⁸)/(1 - -2) = -3×(1-256)/3 = -3×-255/3 =

= -1×-255 = 255

4 0
3 years ago
Share 747 pound in the ratio 2:7 betwen tom and ben
Yuliya22 [10]
2p + 7p = 747
9p = 747 divded by 9 = 83
2 * 83 = 166    7 * 83 = 581

Hope it helps
7 0
3 years ago
(x^2–x–4)multiplied by(x–5)
Sergio [31]

(x-5)(x^2-x-4) =x^3-6x^2+x+20

Step-by-step explanation:

Given polynomials are:

x^2-x-4\\and\\x-5

Both polynomials have to be multiplied. We will use distributive property for the said purpose

So,

(x-5)(x^2-x-4)\\=x(x^2-x-4)-5(x^2-x-4)\\=x^3-x^2-4x-5x^2+5x+20

Combining alike terms

=x^3-x^2-5x^2-4x+5x+20\\=x^3-6x^2+x+20

Hence,

(x-5)(x^2-x-4) =x^3-6x^2+x+20

Keywords: Polynomials, Multiplication

Learn more about polynomials at:

  • brainly.com/question/4703807
  • brainly.com/question/4703820

#LearnwithBrainly

7 0
3 years ago
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