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Gekata [30.6K]
3 years ago
8

Which figure has two pairs of congruent sides and more than two lines of symmetry?

Mathematics
2 answers:
Advocard [28]3 years ago
5 0
C. trapezoid

a trapezoid's legs are congruent, while the bases are symmetrical

a square is the answer you would want, because it is always true, while a trapezoid is only "sometimes" true

thanks Kaikers :D


san4es73 [151]3 years ago
5 0
The correct answer is "B" square. A square has 4 equal sides (and angles), so it is the only polygon that has more than 2 lines of symmetry. 

I hope this helps!
~kaikers
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9/200. divide 45 and 1000 both by 5 to reduce it.
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2 years ago
Two step Equations <br> -2x-3=5
tatyana61 [14]

Answer:

X = -4

Step-by-step explanation:

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3 years ago
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Students who party before an exam are twice as likely to fail as those who don't party (and presumably study). If 20% of the stu
True [87]

Answer:

The fraction of the students who failed to went partying = \frac{1}{10}

Step-by-step explanation:

Let total number of students = 100

No. of students partied are twice the no. of students who not partied.

⇒ No. of students partied = 2 × the no. of students who are not partied

No. of students partied before the exam = 20 % of total students

⇒ No. of students partied before the exam = \frac{20}{100} × 100

⇒ No. of students partied before the exam =  20

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8 0
3 years ago
Consider a rabbit population​ P(t) satisfying the logistic equation StartFraction dP Over dt EndFraction equals aP minus bP squa
maria [59]

Solution:

Given :

$\frac{dP}{dt}= aP-bP^2$         .............(1)

where, B = aP = birth rate

            D = $bP^2$  =  death rate

Now initial population at t = 0, we have

$P_0$ = 220 ,  $B_0$ = 9 ,  $D_0$ = 15

Now equation (1) can be written as :

$ \frac{dP}{dt}=P(a-bP)$

$\frac{dP}{dt}=bP(\frac{a}{b}-P)$    .................(2)

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$\frac{dP}{dt}=kP(M-P)$

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This gives us M = a/b

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$a_0=\frac{B_0}{P_0}$    and     $b_0=\frac{D_0}{P_0^2}$

So, $M=\frac{B_0P_0}{D_0}$

          = $\frac{9 \times 220}{15}$

          = 132

Now from equation (2), we get the constants

k = b = $\frac{D_0}{P_0^2} = \frac{15}{220^2}$

        = $\frac{3}{9680}$

The population P(t) from logistic equation is calculated by :

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$P(t)= \frac{29040}{220-88e^{-\frac{396}{9680} t}}$

As per question, P(t) = 110% of M

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$ 220-88e^{\frac{-99}{2420} t}=200$

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Now taking natural logs on both the sides we get

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Answer:

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Step-by-step explanation:

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2 years ago
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