Factor the trinomial: 3x^2 + 20x +25

. Un turist a parcurs un traseu în trei zile. În prima zi turistul a parcurs

Dahasolnce [82]

Answer:

The total length of route travelled in 3 days = 30 km

Lungimea traseului parcurs în cele trei zile = 30 km

Step-by-step explanation:

English Translation of the Question

A tourist traveled a route in three days. On the first day the tourist covered 40% of the length of the route, on the second day the tourist covered 5/6 of the remaining distance after the first day, and on the third day the remaining 3 km. Calculate the length of the route traveled in the three days.

Let the total length of the route be x.

Lungimea traseului parcurs în cele trei zile = x

- On the first day, the tourist covers 40% of the total length of the route;

În prima zi turistul a parcurs 40% din lungimea traseului

That is, 40% of x = 0.4x

- On the second day, the tourist covered 5/6 of the remaining journey.

în a doua zi turistul a parcurs 5/6 din distanța rămasă de parcurs după prima zi

The remaining journey after day 1 = x - 0.4x = 0.6x

(5/6) of the remaining journey = (5/6) × 0.6x = 0.5x

- On the third day, the toursit travels the total remaining length of the route = 3 km

iar în a treia zi restul de 3 km

The total remaining length of the route = x - 0.4x - 0.5x = 0.1x

0.1x = 3

x = (3/0.1) = 30 km

The Hope this Helps!!!

3 0

3 years ago

What is the angle measurement of angle DBC?<br> Help please

Andreyy89

45 _____________________\_

6 0

2 years ago

In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In ad

zysi [14]

Answer: a) 0.0792 b) 0.264

Step-by-step explanation:

Let Event D = Families own a dog .

Event C = families own a cat .

Given : Probability that families own a dog : P(D)=0.36

Probability that families own a dog also own a cat : P(C|D)=0.22

Probability that families own a cat : P(C)= 0.30

a) Formula to find conditional probability :

$P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow P(A\cap B)=P(B|A)\times P(A)$ (1)

Similarly ,

$P(C\cap D)=P(C|D)\times P(D)\\\\=0.22\times0.36=0.0792$

Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792

b) Again, using (2)

$P(D|C)=\dfrac{P(C\cap D)}{P(C)}\\\\=\dfrac{0.0792}{0.30}=0.264$

Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264

5 0

2 years ago

Geometry question for homework.

mars1129 [50]

So remember that the distance formula is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

1.

$\sqrt{(4-4)^2+(-2-(-5))^2}\\ \sqrt{0^2+3^2}\\ \sqrt{0+9}\\ \sqrt{9}\\ 3$

Distance between (4,-5) and (4,-2) is 3 units.

2.

$\sqrt{(1-(-1))^2+(1-4)^2}\\ \sqrt{2^2+(-3)^2}\\ \sqrt{4+9}\\ \sqrt{13}$

Distance between (1,1) and (-1,4) is √13 (or 3.61 rounded to the hundreths) units.

3 0

2 years ago

Select the equivalent expression.

ZanzabumX [31]

the equivalent expression is C

3 0

3 years ago