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2 years ago

Which triangle congruence postulates can be used to prove that Triangle ABD is congruent to Triangle CDB in Rectangle ABCD?

Mathematics

1 answer:

kramer2 years ago

4 0

Answer:

SSS, SAS, ASA, AAS, HL

Step-by-step explanation:

1. SSS (side side side) says if 3 sides of one triangle are congruent to 3 sides of another triangle, then the 2 triangles are congruent.

2. SAS (side angle side) says if 2 sides and the included angle of a triangle are congruent to 2 sides and the included angle of another triangle, then the 2 triangles are congruent.

3. ASA (angle side angle) says if 2 angles and the included side of a triangle are congruent to 2 angles and the included side of another triangle, then the 2 triangles are congruent.

4. AAS (angle angle side) says if 2 angles and the none included side of one triangle are congruent to the corresponding parts of another triangle, then the 2 triangles are congruent.

5 HL (hypotenuse leg) says if 2 right triangles that have a congruent hypotenuse and a corresponding congruent leg, then the 2 triangles are congruent.

Because it is a rectangle, the sides are equal, and they share the same hypotenuse.

You might be interested in

Given that A, O & B lie on a straight line segment, evaluate obtuse ∠AOC.

Evgesh-ka [11]

Answer:

AOC = 124°

Step-by-step explanation:

Angle on a straight line is 180°, therefore, the sum of the three angles shown is 180;

This can be written like so:

(3x + 94) + (x + 30) + (2x - 4) = 180

This equation can be solved to find x:

6x + 120 = 180

6x = 180 - 120

6x = 60

x = 10

AOC = 3(10) + 94

= 30 + 94

= 124

7 0

2 years ago

Read 2 more answers

A square has a length of 4.8 feet. If the square is dilated by a factor of 4 , what is the length of a side of the new square? W

oksano4ka [1.4K]

Answer:

4.8x4=19.2, and then 19.2/4=4.8 feet

Each side would be 4.8 feet

8 0

3 years ago

Select the values that make the inequality y≥2y≥2 true.

ExtremeBDS [4]

9514 1404 393

Answer:

{2, 2.001, 2.01, 2.1, 3, 5, 7, 10}

Step-by-step explanation:

The symbol ≥ means "greater than or equal to", so any values 2 or greater will make the inequality true. From your list, those are ...

2 2.001 2.01 2.1 3 5 7 10

3 0

2 years ago

a washer and dryer cost a total of $936. The cost of the washer is two times the cost of the dryer. find the cost of each item

slamgirl [31]

Step-by-step explanation:

The washer costs $624 and the dryer costs 312.

624 is double 312 so that works out. Also if you add 624 and 312 you get 936.

7 0

2 years ago

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β =

stich3 [128]

I'm assuming $\alpha$ is the shape parameter and $\beta$ is the scale parameter. Then the PDF is

$f_X(x)=\begin{cases}\dfrac29xe^{-x^2/9}&\text{for }x\ge0\\\\0&\text{otherwise}\end{cases}$

a. The expectation is

$E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\frac29\int_0^\infty x^2e^{-x^2/9}\,\mathrm dx$

To compute this integral, recall the definition of the Gamma function,

$\Gamma(x)=\displaystyle\int_0^\infty t^{x-1}e^{-t}\,\mathrm dt$

For this particular integral, first integrate by parts, taking

$u=x\implies\mathrm du=\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X]=\displaystyle-xe^{-x^2/9}\bigg|_0^\infty+\int_0^\infty e^{-x^2/9}\,\mathrm x$

$E[X]=\displaystyle\int_0^\infty e^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ , so that $\mathrm dx=\dfrac32y^{-1/2}\,\mathrm dy$ :

$E[X]=\displaystyle\frac32\int_0^\infty y^{-1/2}e^{-y}\,\mathrm dy$

$\boxed{E[X]=\dfrac32\Gamma\left(\dfrac12\right)=\dfrac{3\sqrt\pi}2\approx2.659}$

The variance is

$\mathrm{Var}[X]=E[(X-E[X])^2]=E[X^2-2XE[X]+E[X]^2]=E[X^2]-E[X]^2$

The second moment is

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2f_X(x)\,\mathrm dx=\frac29\int_0^\infty x^3e^{-x^2/9}\,\mathrm dx$

Integrate by parts, taking

$u=x^2\implies\mathrm du=2x\,\mathrm dx$

$\mathrm dv=xe^{-x^2/9}\,\mathrm dx\implies v=-\dfrac92e^{-x^2/9}$

$E[X^2]=\displaystyle-x^2e^{-x^2/9}\bigg|_0^\infty+2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

$E[X^2]=\displaystyle2\int_0^\infty xe^{-x^2/9}\,\mathrm dx$

Substitute $x=3y^{1/2}$ again to get

$E[X^2]=\displaystyle9\int_0^\infty e^{-y}\,\mathrm dy=9$

Then the variance is

$\mathrm{Var}[X]=9-E[X]^2$

$\boxed{\mathrm{Var}[X]=9-\dfrac94\pi\approx1.931}$

b. The probability that $X\le3$ is

$P(X\le 3)=\displaystyle\int_{-\infty}^3f_X(x)\,\mathrm dx=\frac29\int_0^3xe^{-x^2/9}\,\mathrm dx$

which can be handled with the same substitution used in part (a). We get

$\boxed{P(X\le 3)=\dfrac{e-1}e\approx0.632}$

c. Same procedure as in (b). We have

$P(1\le X\le3)=P(X\le3)-P(X\le1)$

and

$P(X\le1)=\displaystyle\int_{-\infty}^1f_X(x)\,\mathrm dx=\frac29\int_0^1xe^{-x^2/9}\,\mathrm dx=\frac{e^{1/9}-1}{e^{1/9}}$

Then

$\boxed{P(1\le X\le3)=\dfrac{e^{8/9}-1}e\approx0.527}$

7 0

2 years ago