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2 years ago

If CDEF ≅ MNPQ, then ≅ DE ≅

Mathematics

2 answers:

Yanka [14]2 years ago

5 0

The answer is A.NP.

Inga [223]2 years ago

3 0

Answer:

The correct option is a.

Step-by-step explanation:

It is given that CDEF ≅ MNPQ.

We need to find the side of MNPQ which is congruent to the side DE.

The corresponding parts of congruent figures are congruent.

Since CDEF ≅ MNPQ, therefore the corresponding sides of both figures are congruent.

$CD\cong MN$

$DE\cong NP$

$EF\cong PQ$

$CF\cong MQ$

The side NP is congruent to side DE. Therefore the correct option is a.

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Mopeds (small motorcycles with an engine capacity below 50 cm3) are very popular in Europe because of their mobility, ease of op

Alex17521 [72]

Answer:

a) $P(X$

And we can find this probability using the normal standard table or excel and we got:

$P(z$

b) $P(X>48)=P(\frac{X-\mu}{\sigma}>\frac{48-\mu}{\sigma})=P(Z>\frac{48-46.8}{1.75})=P(z>0.686)$

And we can find this probability using the complement rule, the normal standard table or excel and we got:

$P(z>0.686)=1- P(Z$

c) $P(46.8-1.5*1.75$

And we can find this probability with this difference:

$P(-1.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-1.5$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the velocities of a population, and for this case we know the distribution for X is given by:

$X \sim N(46.8,1.75)$

Where $\mu=46.8$ and $\sigma=1.75$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using the normal standard table or excel and we got:

$P(z$

Part b

$P(X>48)=P(\frac{X-\mu}{\sigma}>\frac{48-\mu}{\sigma})=P(Z>\frac{48-46.8}{1.75})=P(z>0.686)$

And we can find this probability using the complement rule, the normal standard table or excel and we got:

$P(z>0.686)=1- P(Z$

Part c

$P(46.8-1.5*1.75$

And we can find this probability with this difference:

$P(-1.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-1.5$

6 0

2 years ago

What is the logarithm of the equilibrium constant, log K, at 25°C of the voltaic cell constructed from the following two half-re

sashaice [31]

Answer:

4.0921 is the logarithm of the equilibrium constant.

Step-by-step explanation:

$Fe^{2+} (aq) +2e^{-}\rightarrow Fe(s)$ ; E° = - 0.41 V

$Ag^+(aq) + e^-\rightarrow Ag(s)$ ; E° = 0.80 V

Iron having negative value of reduction potential .So ,that means that it will loose electron easily and get oxidized.Hence, will be at anode.

$E^{o}_{cell}$ =Reduction potential of cathode - Reduction potential of anode

$E^{o}_{cell}=E^{o}_c-E^{o}_a$

$=0.80 V-(-0.41 V)=1.21 V$

$Fe^{2+} (aq) + 2e^{-}\rightarrow Fe(s)$ ; E° = - 0.41 V

$2Ag^+(aq) + 2e^-\rightarrow 2Ag(s)$ ; E° = 0.80 V

Net reaction: $Fe(s)+2Ag^{+}\rightarrow Fe^{2+}+2Ag(s)$

n = 2

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 1.21 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

Putting values in above equation, we get:

$2\times 96500\times 1.21 V=8.314\times 298\times \ln K_{eq}$

$\ln K_{eq}=9.3478$

$\log K_{eq}=\frac{9.3478}{2.303}=4.0921$

4.0921 is the logarithm of the equilibrium constant.

7 0

2 years ago

Wynona buys some new boots that are priced $85.79. She gets successive discounts of 10% followed by 5% off on the boots. The pur

ddd [48]

the correct answer is D

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2 years ago

Read 2 more answers

Sue scored a total of 35 points in two games.she scored 6 times as many points in the second game than in the first.how many mor

kirill115 [55]

Okay,
Both games are: 35
First game: ?
Second game: 6 more than the first.
So,
first we subtract 6 from 35.
35 - 6 = 29
Divide by 2.
49 divided by 2 = 14.5
Add the 6 point= 14.5 + 6 = 20.5
To make sure add.
First game: 14.5
Second game:20.5
14.5 + 20.5 = 35

3 0

2 years ago

Solve the following addition and subtraction problems. a. 3 km 9 hm 9 dam 19 m + 7 km 7 dam b. 5 sq.km 95 ha 8,994 sq.m + 11 sq.

kipiarov [429]

As a general rule to solve the problem we are going to transform all values to the lower unit.
a. 3 km 9 hm 9 dam 19 m + 7 km 7 dam
3,000 m 900 m 90 m 19 m + 7,000 m 70 m = 4,009 + 7,070 = 11,079 m
b. 5 sq.km 95 ha 8,994 sq.m + 11 sq. km. 11 ha 9,010 sq. m.
5,000,000 sq m 95,0000 sq m 8,994 sq m + 11,000,000 sq m 110,000 sq
9,010 sq m
5,103,994 sq m + 11,119,010 sq m = 16,223,004 sq m
c. 44 m â€“ 5 dm
44 m - 0.5 m = 43.5 m
d. 73 km 47 hm 2 dam - 11 km 55 hm

73,000 m 4,700 m 20 m - 11,000 m 5,500 m
77,720 m - 16,500 m = 61,220 m

5 0

2 years ago