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Anton [14]
3 years ago
10

Which criteria for triangle congruence can be used to prove the pair of triangles below are congruent?

Mathematics
2 answers:
Firlakuza [10]3 years ago
8 0

Answer: The correct option is fourth, i.e., SSS.

Explanation:

First write the name of vertices as shown below,

From the figure we can say that in triangle DAB and triangle BCD,

AB=CD

DA=BC

In both triangle the sides BD is common, so,

BD=BD

Since three sides of triangles DAB and triangle BCD are same so by the SSS property of congruent triangle we can conclude that the triangle DAB and triangle BCD are congruent.

Therefore the fourth option is correct.

kakasveta [241]3 years ago
4 0

Answer:

SSS

Step-by-step explanation:

We are given that two triangles.

We have to find the criteria can be used to prove the pair of triangles are congruent.

In given pair of triangles,

Two sides of one triangle are equal to corresponding two sides of other triangle.

One side is common in both triangles.

When three sides of one triangle are equal to three sides of other triangle then the triangles are congruent by SSS postulate.

Hence, the pair of triangles are congruent by SSS postulate.

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Answer:

The recursive formula is:

Cn = 10C(n-1)

Step-by-step explanation:

Given the geometric sequence.

-0.56, -5.6, -56, -560, ...

The common ratio is

-5.6/-0.56 = -56/-5.6 = -560/-56 = ... = 10

The recursive formula is easily

Cn = C(n-1) × 10

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Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and
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Answer:

(0,0)   (4000,0) and (500,79)

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW

\frac{dW}{dt} = -0.02W + 0.00004RW

To solve this, we first equate \frac{dR}{dt} and \frac{dW}{dt} to 0.

So, we have:

0.09R(1 - 0.00025R) - 0.001RW = 0

-0.02W + 0.00004RW = 0

Factor out R in 0.09R(1 - 0.00025R) - 0.001RW = 0

R(0.09(1 - 0.00025R) - 0.001W) = 0

Split

R = 0   or 0.09(1 - 0.00025R) - 0.001W = 0

R = 0   or  0.09 - 2.25 * 10^{-5}R - 0.001W = 0

Factor out W in -0.02W + 0.00004RW = 0

W(-0.02 + 0.00004R) = 0

Split

W = 0 or -0.02 + 0.00004R = 0

Solve for R

-0.02 + 0.00004R = 0

0.00004R = 0.02

Make R the subject

R = \frac{0.02}{0.00004}

R = 500

When R = 500, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0

0.09 -0.01125 - 0.001W = 0

0.07875 - 0.001W = 0

Collect like terms

- 0.001W = -0.07875

Solve for W

W = \frac{-0.07875}{ - 0.001}

W = 78.75

W \approx 79

(R,W) \to (500,79)

When W = 0, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0

0.09 - 2.25 * 10^{-5}R = 0

Collect like terms

- 2.25 * 10^{-5}R = -0.09

Solve for R

R = \frac{-0.09}{- 2.25 * 10^{-5}}

R = 4000

So, we have:

(R,W) \to (4000,0)

When R =0, we have:

-0.02W + 0.00004RW = 0

-0.02W + 0.00004W*0 = 0

-0.02W + 0 = 0

-0.02W = 0

W=0

So, we have:

(R,W) \to (0,0)

Hence, the points of equilibrium are:

(0,0)   (4000,0) and (500,79)

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