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3 years ago

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triangle ABC is shown in the diagram. The links of the sides are in terms of the variable n, where n >4. complete the inquali

ty

2 answers:

drek231 [11]3 years ago

6 0

Answer:

M- B M-A M-C

Step-by-step explanation:

yw

Blababa [14]3 years ago

5 0

9514 1404 393

Answer:

m∠B < m∠A < m∠C

Step-by-step explanation:

We can work with the triangle inequality to find that the side measures form a triangle when n > 5/4. For the given value of n ≥ 4, we don't need to be concerned with whether a triangle is formed or not.

For n = 4, the side lengths are ...

a = 2(4) = 8

b = (4) +3 = 7

c = 3(4) -2 = 10

The longest side is opposite the largest angle, so the ordering of angles is ...

m∠B < m∠A < m∠C

_____

The triangle inequality requires all of these inequalities be true:

a+b > c ⇒ 3n+3 > 3n-2 . . . always true
b+c > a ⇒ 4n+1 > 2n ⇒ n > -1/2
c+a > b ⇒ 5n-2 > n+3 ⇒ n > 5/4

That will be the case for n > 5/4. The attached graph shows the sides and angles keep the same order for n > 3.

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In a quadrilateral ABCD, AB ∥ DC and AD ∥ BC . Find the perimeter of ΔCOD if the diagonals of the quadrilateral intersect each o

SVEN [57.7K]

Answer Perimeter = 33in

Step-by-step explanation:

AC = 20

BD = 20

AB = 13

O is the midpoint

OD = OC = 20/2 = 10

AB = DC = 13 in

P = DC + OD + OC = 13+10+10 = 33in

5 0

3 years ago

Help it’s for a quiz

Vilka [71]

Answer:

i cant answer that

Step-by-step explanation: I cant answer because if it was for a quiz i bet the quiz is over now so there is no point in answering.

6 0

2 years ago

PLSSS HELP ME thanks :))

krok68 [10]

that's all , all the best

4 0

3 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

8 0

3 years ago

Suppose r(t)=costi+sintj+3tk represents the position of a particle on a helix, where z is the height of the particle above the g

Ilia_Sergeevich [38]

a. The $\vec k$ component tells you the particle's height:

$3t=16\implies t=\dfrac{16}3$

b. The particle's velocity is obtained by differentiating its position function:

$\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=-\sin t\,\vec\imath+\cos t\,\vec\jmath+3\,\vec k$

so that its velocity at time $t=\frac{16}3$ is

$\vec v\left(\dfrac{16}3\right)=-\sin\dfrac{16}3\,\vec\imath+\cos\dfrac{16}3\,\vec\jmath+3\,\vec k$

c. The tangent to $\vec r(t)$ at $t=\frac{16}3$ is

$\vec T(t)=\vec r\left(\dfrac{16}3\right)+\vec v\left(\dfrac{16}3\right)t$

4 0

3 years ago