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Novosadov [1.4K]
2 years ago
4

Right triangle ABC is reflected over AC, then dilated by a scale factor of Two-thirds to form triangle DEC. Which statements abo

ut the two triangles must be true? Select three options. △ABC ~ △DEC ∠B ≅ ∠E 3BC = 2EC 3DE = 2AB 3m∠A = 2m∠D 2m∠A = 3m∠D

Mathematics
2 answers:
anzhelika [568]2 years ago
7 0

Answer:

1

2

3

Step-by-step explanation:

hichkok12 [17]2 years ago
4 0

Answer:

\triangle ABC\sim \triangle DEC\\ \\\angle B\cong \angle E\\ \\3BC=2EC

Step-by-step explanation:

Right triangle ABC is reflected over AC to form triangle AFC.

Triangle AFC dilated by a scale factor of two-thirds to form triangle DEC.

Reflection is a rigid transformation which preserves lengths and angles, then

\triangle ABC\cong \triangle AFC

Dilation is a transformation which maps the triangle into similar triangle, so

\triangle AFC \sim \triangle DEC

Thus,

\triangle ABC \sim \triangle DEC

Similar triangle have congruent corresponding angles, so

\angle B\cong \angle E

and proportional corresponding sides

\dfrac{BC}{EC}=\dfrac{AB}{DE}=\dfrac{2}{3}\Rightarrow \\ \\3BC=2EC\\ \\3AB=2DE

Thus, correct options are:

\triangle ABC\sim \triangle DEC\\ \\\angle B\cong \angle E\\ \\3BC=2EC

You might be interested in
Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C cont
castortr0y [4]

Answer:

k= 80%

Step-by-step explanation:

Jar A contains 4*0.45 L acid, and 4 L of a solution  of acid.

Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.

Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.

50% = 0.5

For jar A.

(2/3)*k/100 L acid  is added to jar A.

Now jar A contains   4*0.45 L + (2/3)*k/100 L acid, and it has (4+2/3)L of a solution.

L solute/L solution = 0.5

[4*0.45 L + (2/3)*k/100 L]/(4+2/3)L = 0.5

[1.8 + (2k/300)]/[(12+2)/3] = 0.5

[1.8 + (2k/300)]/[14/3] = 0.5

[1.8 + (2k/300)]= 0.5*(14/3)

(2k/300) = 0.5*(14/3) - 1.8

2k = (0.5*(14/3) - 1.8)*300

k = (0.5*(14/3) - 1.8)*300/2 =80

k= 80%

We also can find k using jar B.

(1/3)k/100 L acid is added  to jar B.

Now jar B contains 5*0.48 L+ (1/3)k/100 L acid, and it has (5+1/3) L of a solution.

L solute/L solution = 0.5

[5*0.48 L+ (1/3)k/100 L ]/(5+1/3)L= 0.5

[5*0.48 + (1/3)k/100 ]/(5+1/3)= 0.5

This equation also gives k=80%

Check.

We can check at least for jar A.

Jar A has 4L solution and 4*0.45=1.8 L acid.

2/3 L of the solution from jar C was added, and now we have 4 2/3 L of solution.

(2/3)* 80%= (2/3)*0.8 acid was added from jar C.

Now we have [1.8 +(2/3)*0.8] L acid in jar A.

L solute/L solution =  [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50%  as it is given that jar A has 50% at the end.

7 0
2 years ago
Suppose X is the breaking strength (newtons) of a material, and X is normally distributed with
Veronika [31]

Answer:

a) 0.997 is the  probability that the breaking strength is at least 772 newtons.

b) 0.974  is the probability that this material has a breaking strength of at least 772 but not more  than 820    

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 800 newtons

Standard Deviation, σ = 10 newtons

We are given that the distribution of  breaking strength is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P( breaking strength of at least 772 newtons)

P(x \geq 772)

P( x \geq 772) = P( z \geq \displaystyle\frac{772 - 800}{10}) = P(z \geq -2.8)

= 1 - P(z

Calculation the value from standard normal z table, we have,  

P(x \geq 772) = 1 - 0.003 = 0.997 = 99.7\%

0.997 is the  probability that the breaking strength is at least 772 newtons.

b) P( breaking strength of at least 772 but not more  than 820)

P(772 \leq x \leq 820) = P(\displaystyle\frac{772 - 800}{10} \leq z \leq \displaystyle\frac{820-800}{10}) = P(-2.8 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2.8)\\= 0.977 - 0.003 = 0.974 = 97.4\%

P(772 \leq x \leq 820) = 97.4\\%

0.974  is the probability that this material has a breaking strength of at least 772 but not more  than 820.

7 0
2 years ago
G find the area of the surface over the given region. use a computer algebra system to verify your results. the sphere r(u,v) =
Svetach [21]
Presumably you should be doing this using calculus methods, namely computing the surface integral along \mathbf r(u,v).

But since \mathbf r(u,v) describes a sphere, we can simply recall that the surface area of a sphere of radius a is 4\pi a^2.

In calculus terms, we would first find an expression for the surface element, which is given by

\displaystyle\iint_S\mathrm dS=\iint_S\left\|\frac{\partial\mathbf r}{\partial u}\times\frac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\dfrac{\partial\mathbf r}{\partial u}=a\cos u\cos v\,\mathbf i+a\cos u\sin v\,\mathbf j-a\sin u\,\mathbf k
\dfrac{\partial\mathbf r}{\partial v}=-a\sin u\sin v\,\mathbf i+a\sin u\cos v\,\mathbf j
\implies\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}=a^2\sin^2u\cos v\,\mathbf i+a^2\sin^2u\sin v\,\mathbf j+a^2\sin u\cos u\,\mathbf k
\implies\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|=a^2\sin u

So the area of the surface is

\displaystyle\iint_S\mathrm dS=\int_{u=0}^{u=\pi}\int_{v=0}^{v=2\pi}a^2\sin u\,\mathrm dv\,\mathrm du=2\pi a^2\int_{u=0}^{u=\pi}\sin u
=-2\pi a^2(\cos\pi-\cos 0)
=-2\pi a^2(-1-1)
=4\pi a^2

as expected.
6 0
2 years ago
Rewrite in simplest radical form! Show your work.
vazorg [7]

x^{\frac{1}{\frac{-3}{6} }}

First, let's deal with the fraction in the denominator of the exponent. Multiply the top and bottom of the exponent by 6.

x^{\frac{6}{-3} }

Now that the fraction in the denominator is taken care of, we can reduce the denominator.

x^{-2}. Some professors might accept this as simplest form, but others might ask you to get rid of the negative.

x^{-2} = \frac{1}{x^{2} }

7 0
2 years ago
Read 2 more answers
Julie wants to purchase a new home in six years.
sergejj [24]

Good for Julie!  Is there a question that comes with this post?

-TTL

8 0
2 years ago
Read 2 more answers
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