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dalvyx [7]
2 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]2 years ago
7 0
Your anwser for your question is QP Qr and Qrs
AysviL [449]2 years ago
4 0

Answer:

QP = QR

PM = MR

QRS = QPS

Step-by-step explanation:

took the test haha

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In algebra, we often study relationships where a change to one variable causes change in another variable. Describe a situation
dexar [7]

Answer:

The amount of firewood we use gives us a handful of ash. SO, on the x axes would we the amount of firewood and on the y axis it would be the amount of ash that was concluded.

Step-by-step explanation:

I have this q rn. And this is what im using. hope this helps!

4 0
2 years ago
Origanal 240 pages 20% increase
FrozenT [24]
10 % = 24
20 % = 2 x 10%
20 % = 48
240 + 48 = 288

hope this helps you
6 0
3 years ago
Read 2 more answers
Find p(2) and p(4) for the function p(x) = 6x4 + 4x3 – 3x2 + 8x + 15.
gulaghasi [49]

Answer:

p(2) =147 and p(4) = 1791

Step-by-step explanation:

We are given p(x)= 6x^4 + 4x^3 – 3x^2 + 8x + 15.

Now we need to find value of p(2) and p(4)

Put x =2,

p(2) = 6(2)^4 + 4(2)^3 – 3(2)^2 + 8(2) + 15

p(2) = 6(16)+4(8)-3(4)+8(2)+15

p(2) = 96+32-12+16+15

p(2) = 147

Now put x = 4

p(4) = 6(4)^4 + 4(4)^3 – 3(4)^2 + 8(4) + 15

p(4) = 6(256)+4(64)-3(16)+8(4)+15

p(4) = 1536+256-48+32+15

p(4) = 1791

3 0
3 years ago
Read 2 more answers
Please help me out thank you
Leno4ka [110]

Answer:

  1. (-0.0008)^0 =1
  2. 1291^1 =1
  3. (1/3)^0 = 1

Step-by-step explanation:

\mathrm{Apply\:exponent\:rule}:\quad \:a^0=1,\:\\\\(-0.0008)^0 = 1\\\\1291^0 =1\\\\-(-3141)^0 = -1\\\\-(\frac{5}{8} )^0 =-1\\\\0^1 = 0\\\\(\frac{1}{3} )^0 = 1\\

6 0
3 years ago
How do I find a2 and a3 for the following geometric sequence? 54, a2, a3, 128
vlada-n [284]
The formula for the nth term of a geometric sequence:
a_n=a_1 \times r^{n-1}
a₁ - the first term, r - the common ratio

54, a_2, a_3, 128 \\ \\
a_1=54 \\
a_4=128 \\ \\
a_n=a_1 \times r^{n-1} \\
a_4=a_1 \times r^3 \\
128=54 \times r^3 \\
\frac{128}{54}=r^3 \\ \frac{128 \div 2}{54 \div 2}=r^3 \\
\frac{64}{27}=r^3 \\
\sqrt[3]{\frac{64}{27}}=\sqrt[3]{r^3} \\
\frac{\sqrt[3]{64}}{\sqrt[3]{27}}=r \\
r=\frac{4}{3}

a_2=a_1 \times r= 54 \times \frac{4}{3}=18 \times 4=72 \\
a_3=a_2 \times r=72 \times \frac{4}{3}=24 \times 4=96 \\ \\
\boxed{a_2=72, a_3=96}
7 0
3 years ago
Read 2 more answers
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