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melomori [17]
1 year ago
14

Which piece of additional information can be used to prove that △rst ~ △vut? rt = st 3st = ut ∠r ≅ ∠v ∠v ≅ ∠u

Mathematics
1 answer:
Sunny_sXe [5.5K]1 year ago
3 0

The option third ∠R ≅ ∠V is correct because line segment RS and UV are parallel which is crossed by a transversal RV.

<h3>What is the similarity law for triangles?</h3>

It is defined as the law to prove that the two triangles have the same shape, but it is not compulsory to have the same size. The ratio of the corresponding sides is in the same proportions and the corresponding angles are congruent.

The question is incomplete.

The complete question is in the picture, please refer to the attached picture.

We have given two triangles:

Triangle RST and Triangle VUT

AS we can see Line segment RS and UV are parallel.

Which is crossed by a transversal RV.

The angle SRV = Angle RVU

Thus, the option third ∠R ≅ ∠V is correct because line segment RS and UV are parallel which is crossed by a transversal RV.

Learn more about the similarity of triangles here:

brainly.com/question/8045819

#SPJ1

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Which angle has a sine of -1/2 and a cosine of -√3/2
Sergeeva-Olga [200]
\sin x=-\dfrac{1}{2} < 0\\\\\cos x=-\dfrac{\sqrt3}{2} < 0\\\\therefore\ x\in(180^o;\ 270^o)

\text{We know:}\ \tan x=\dfrac{\sin x}{\cos x}\\\\\text{therefore:}\ \tan x=\dfrac{-\frac{1}{2}}{-\frac{\sqrt3}{2}}=\dfrac{1}{2}\cdot\dfrac{2}{\sqrt3}=\dfrac{1}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{\sqrt3}{3}

\tan x=\dfrac{\sqrt3}{3}\Rightarrow x=30^o+180^o\cdot k;\ k\in\mathbb{Z}

x\in(180^o;\ 270^o)\ therefore\ \boxed{x=30^o+180^o=210^o}
8 0
3 years ago
Just tell me the answer and how you got it , thank you​
leva [86]

(2x^2 + 4x + 6) - (-3x - 5x^2 + 4)

The negative sign separating the expressions is just multiplying the rightmost expression by -1.

(2x^2 + 4x + 6) + (3x + 5x^2 - 4)

Now, let's add together terms that share the same variable.

2x^2 + 5x^2 = 7x^2

4x + 3x = 7x

6 + (-4) = 2

Let's combine each of these terms into one expression.

7x^2 + 7x + 2 is the final answer.

7 0
3 years ago
Please help I don't understand!
vovikov84 [41]

\large{\underline{\underline{\pmb{\frak {\color {blue}{Answer:}}}}}}

True

\sf \green{Explanation}

As it is said in the question the probability of getting a head is 0.5. And, if we toss it two times it will be 0.25.

We'll have to multiply 0.5 with 0.5.

\boxed{ \frak \red{brainlysamurai}}

8 0
1 year ago
A+b+c=4<br> a²+b²+c²=10<br> a³+b³+c³=22<br> a4+b4+c²=?
iragen [17]

Answer:

34

Step-by-step explanation:

6 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%20%5Crm%20%20%5Clim_%7Bk%20%5Cto%20%5Cinfty%20%7D%20%5Csqrt%5B%20%20k%5D%7B%20%5CGamma%20%
Naddik [55]

We have

\sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)\right)}k\right) \\\\ = \exp\left(\dfrac{\ln\left(\Gamma\left(\dfrac1k\right)\right)+\ln\left( \Gamma\left(\dfrac2k\right)\right)+ \cdots +\ln\left(\Gamma\left(\dfrac kk\right)\right)}k\right)

and as k goes to ∞, the exponent converges to a definite integral. So the limit is

\displaystyle \lim_{k\to\infty} \sqrt[k]{\Gamma\left(\dfrac1k\right) \Gamma\left(\dfrac2k\right) \cdots \Gamma\left(\dfrac kk\right)} \\\\ = \exp\left(\lim_{k\to\infty} \frac1k \sum_{i=1}^k \ln\left(\Gamma\left(\frac ik\right)\right)\right) \\\\ = \exp\left(\int_0^1 \ln\left(\Gamma(x)\right)\, dx\right) \\\\ = \exp\left(\dfrac{\ln(2\pi)}2}\right) = \boxed{\sqrt{2\pi}}

6 0
2 years ago
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