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ludmilkaskok [199]
3 years ago
6

Need help ASAP NO LINKS

Mathematics
2 answers:
Stella [2.4K]3 years ago
8 0
Hope this helps! feel free to clarify

matrenka [14]3 years ago
6 0

Answer:

9

Step-by-step explanation:

5(n-2)=35

5n-10=35

5n=35+10

5n=45

n=45/5

n=9

Please mark me as brainliest

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Compute the value of the discriminant and give the number of real solutions to the quadratic equation:
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If the delta is less than zero and if no polynomial real roots
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What number should be added to 77 to make the sum 0?
Lina20 [59]

77 + x = 0

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Let f be a functions of degree 4 whose coefficients are real numbers: two of its zeros are - 3 and 4 - i. Explain why one of the
kvasek [131]

Answer:

Step-by-step explanation:

We have the following theorem, if f is a polynomial with real coefficients, we can factor it completely in factors of the degree at most 2.

Consider first a polynomial of degree two, hence it is a polynomial of the form ax^2+bx+c. The cuadratic formula tells us that the solutions are of the form

x = \frac{-b\pm \sqrt[]{b^2-4ac}}{2a}.

Note that square root, over the reals, tells us that the are only real solutions if b^2-4ac \geq 0. If that is not the case, say it's negative, the solution are complex. Then, the solutions are of the form

x = \frac{-b \pm i \sqrt[]{4ac-b^2}}{2a}. NOte that this means that if we have a complex number of the form a+bi that is a solution, then the number a-bi (who is called the complex conjugate) is also a solution.

Recall that when we have a polynomial f(x) whose a zero is the number c, then we can factor f as follows f(x) = (x-c) * p(x) where p(x) is another polynomial of lesser degree .

So far, we know that -3 and 4-i are zeros of the function f. Note that we are missing two zeros. But, since complex numbers are zeros of polynomial only by pairs (that is the number and its conjugate are zeros), then, we must have that one of the missing 2 zeros is a real number. We have 4-i as a zero, then, its complex conjugate must be also a zero, i.e 4+i is a zero.

8 0
3 years ago
Whats this question im confused 9/250
romanna [79]
Its probably a division question. 250 divided by nine which is 27.7 which rounds to 28.
8 0
3 years ago
Una heladeria dispone de 20 frutas distintas para elaborar sus malteadas. Si los clientes pueden elegir tres sabores para mezcla
Dahasolnce [82]

Answer:

Existen 6840 permitaciones de malteadas de tres sabores distintos que la heladería puede ofrecer.

Step-by-step explanation:

En este caso, el cliente que adquiere una malteada de tres sabores distintos sigue el siguiente procedimiento:

1) El primer sabor sale de cualquiera de las 20 frutas disponibles.

2) El segundo sabor es distinto al primer sabor, es decir, que sale de las 19 frutas restantes.

3) El tercer sabor es distinto al primer sabor y al segundo sabor, es decir, que sale de las 18 frutas restantes.

Puesto que existe una doble conjunción y que puede importar el orden según la preferencia del cliente, se habla matemáticamente de una permutación, definida como:

n\mathbb{P}k = \frac{n!}{(n-k)!} (1)

Donde:

n - Número de sabores disponibles, adimensional.

k - Número de sabores escogidos, adimensional.

Si tenemos que n = 20 y k = 3, entonces la cantidad de malteadas de tres sabores distintos es:

n\mathbb{P}k = \frac{20!}{(20-3)!}

n\mathbb{P}k = \frac{20!}{17!}

n\mathbb{P}k = 20\cdot 19\cdot 18

n\mathbb{P}k = 6840

Existen 6840 permitaciones de malteadas de tres sabores distintos que la heladería puede ofrecer.

5 0
3 years ago
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