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2 years ago

If you multiply two prime numbers, will the result be prime or composite?

Mathematics

1 answer:

ki77a [65]2 years ago

4 0

Answer:

It can never be a prime number.

Step-by-step explanation:

This is because the product of the two prime numbers are divisible by those two numbers, therefore going against the definition of a prime number. For example 3 and 5 are prime numbers and their product is 15. 15 can be divided by 3 and 5 so it is not a prime number.

Hope this helps.

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Which equation has infinite solutions?

AnnyKZ [126]

The equation that has an infinite number of solutions is $2x + 3 = \frac{1}{2}(4x + 2) + 2$

<h3>How to determine the equation?</h3>

An equation that has an infinite number of solutions would be in the form

a = a

This means that both sides of the equation would be the same

Start by simplifying the options

3(x – 1) = x + 2(x + 1) + 1

3x - 3 = x + 3x + 2 + 1

3x - 3 = 4x + 3

Evaluate

x = 6 ----- one solution

x – 4(x + 1) = –3(x + 1) + 1

x - 4x - 4 = -3x - 3 + 1

-3x - 4 = -3x - 2

-4 = -2 ---- no solution

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

2x + 3 = 2x + 1 + 2

2x + 3 = 2x + 3

Subtract 2x

3 = 3 ---- infinite solution

Hence, the equation that has an infinite number of solutions is $2x + 3 = \frac{1}{2}(4x + 2) + 2$

Read more about equations at:

brainly.com/question/15349799

#SPJ1

<u>Complete question</u>

Which equation has infinite solutions?

3(x – 1) = x + 2(x + 1) + 1

x – 4(x + 1) = –3(x + 1) + 1

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

$\frac 13(6x - 3) = 3(x + 1) - x - 2$

5 0

2 years ago

Why is it easy to compare a positive number and a negative number?

Elza [17]

It is easier to compare a positive number and a negative number because it’s just like regular adding

8 0

2 years ago

I just need help finding h pleasww

SCORPION-xisa [38]

Answer:

h = (2/3)√3
b = 3√2

Step-by-step explanation:

It is useful to remember the ratios between the side lengths of these special triangles.

30°-60°-90° ⇒ 1 : √3 : 2

45°-45°-90° ⇒ 1 : 1 : √2

h is the shortest side, and the given length is the intermediate side. This means ...

h/1 = 2/√3

h = 2/√3 = (2/3)√3 . . . . . . simplify, rationalize the denominator

b is the longest side, and the given length is the short side. This means ...

b/√2 = 3/1

b = 3√2 . . . . . multiply by √2

8 0

2 years ago

Which of the following is the solution to the following system of inequalities?

jek_recluse [69]

Answer:

Step-by-step explanation:

for example point (2,0) is one of the solutions.

=> x + 2y <4

2+ 2(0) = 2 => 2< 4 (true)

=> 3x – y > 2

3(2)-0= 6 => 6>2 (true)

so, the option D is the correct one

7 0

2 years ago

Read 2 more answers

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

Maru [420]

Parameterize $S{/tex] by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k$

with $0\le u\le1$ and $0\le v\le1$ .

Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k$

Then the flux of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}$

6 0

3 years ago