1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
umka21 [38]
2 years ago
15

Giving brainliest to whoever can give me the right answers :)

Mathematics
1 answer:
vredina [299]2 years ago
8 0

Answer:

The First problem is empty

The second is (-oo,3) U [6,oo)

Step-by-step explanation:

Set C is defined for all reals less than or equal to. 3.

Set D is defined for all reals greater than or equal to 6.

The First cardinal want us to find when the two Set intersects. They never intersect so that the Set is empty.

The next cardinal want us to find when the Set contains both of their subsets.

Set C can be from any negative number up to 3 inclusive.

Set D start at 6, exclusively and continues forever to thr positve. So

the Set C And Set D will be

(-oo, 3) U [6,oo)

You might be interested in
Si el lado de un cuadrado mide 10cm cuanto mide él área del círculo que se encuentra adentro de ese cuadrado
Alex17521 [72]

Answer:

El área del círculo que se encuentra en el cuadrado es de 78.5cm²

Step-by-step explanation:

Para resolver este ejercicio tenemos que pensar que un cuadrado tiene sus 4 lados iguales, por lo que todos sus lados medirán 10cm.

Ahora nos fijamos que necesitamos saber para calcular el área de un circulo

a = área

r = radio

π = 3.14

a = π * r²

como podemos ver no sabemos el valor del radio

como el circulo toca con los 4 lados del cuadrado sabemos que su radio sera la distancia del centro del cuadrado a cualquiera de los lados.

Entonces tenemos que dividir un lado por 2

10cm/2 = 5cm

El radio del circulo sera 5cm

Ahora que tenemos todos los datos podemos calcular el valor del área

a = 3.14 * (5cm)²

a = 3.14 * 25cm²

a = 78.5cm²

El área del círculo que se encuentra en el cuadrado es de 78.5cm²

8 0
3 years ago
What is the volume of a rectangular prism with side lengths of x, x +4, and 2x-1
marysya [2.9K]

kkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk

6 0
3 years ago
GUYS please
Advocard [28]

Answer:

2.64

Step-by-step explanation:

0.9(x + 1.4) - 2.3 + 0.1x = 1.6

0.9x + 1.26 -2.3 + 0.1x - 1.6 = 0

(0.9x + 0.1x) + (1.26 - 2.3 - 1.6) = 0  

x - 2.64 = 0

x = 2.64

8 0
3 years ago
A box has four cards numbered 1, 2, 3, and 4
lora16 [44]
Sample space: H1, H2, H3, H4, T1, T2, T3, T4
All outcomes if the card is three: H3, T3
I was unclear about the last two parts of the question.
6 0
3 years ago
A right triangle has legs of 5 ft and 6 ft. What is the length of the hypotenuse? _____ ft.
yarga [219]

Answer:

2.) 7.8

Step-by-step explanation:

We can use Pythagorean theorem

a^2 + b^2 = c^2 where a and b are the legs and c is the hypotenuse

5^2 + 6^2 = c^2

25+36 = c^2

61 = c^2

Take the square root of each side

sqrt(61) = sqrt(c^2)

7.810249676 = c

7 0
4 years ago
Read 2 more answers
Other questions:
  • Find two numbers, if their sum is 97 and their difference is 7.
    12·1 answer
  • The rectangle below has an area of x(2 power) -6x -7 square meters and a width of x-7 meters.
    13·1 answer
  • Lim as x approaches 2pi/3 from the right of csc x, solve by substituting csc with sin
    13·1 answer
  • According to some students, what is the true purpose of homework?
    5·1 answer
  • Solve |2x - 5| = 4. {x | x = -4.5 or x = 4.5} {x | x = 0.5 or x = 4.5} {x | 0.5 < x < 4.5}
    10·1 answer
  • Plz do number fifteen please!
    14·1 answer
  • The hypotenuse of a triangle is 37 millimeters long. One of the legs is 24.5 millimeters long. What is the measure of the missin
    12·1 answer
  • Help pls just help i can't take this anymore​
    5·2 answers
  • <img src="https://tex.z-dn.net/?f=12%20%5Cfrac%7B3%7D%7B6%7D%20%20%2B%2014%5Cfrac%7B4%7D%7B6%7D%20" id="TexFormula1" title="12 \
    6·1 answer
  • What is the answer to this?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!