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3 years ago

Can someone please help me with this?

Mathematics

1 answer:

vredina [299]3 years ago

4 0

$\implies {\blue {\boxed {\boxed {\purple {\sf {A. \:-54}}}}}}$

$\large\mathfrak{{\pmb{\underline{\red{Step-by-step\:explanation}}{\red{:}}}}}$

We have,

$'m = 18'$ and $'a = - 3'$

Substituting the values of $m$ and $a$ in $F=ma$ ,

➺ $\:F = 18 \times - 3$

➺ $\: F = - 54$

Therefore, the value of F is $\boxed{-54}$ .

F = force

m = mass

a = acceleration.

$\large\mathfrak{{\pmb{\underline{\orange{Mystique35 }}{\orange{❦}}}}}$

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4 years ago

write a polynomial function of least degree with integral coefficients that has the given zeros. -(1/3), -i

inessss [21]

Answer:

$f(x)=3x^3+x^2+3x+1$

Step-by-step explanation:

If a real number $-\frac{1}{3}$ is a zero of polynomial function, then

$x-\left(-\dfrac{1}{3}\right)=x+\dfrac{1}{3}$

is the factor of this function.

If a complex number $-i$ is a xero of the polynomial function, then the complex number $i$ is also a zero of this function and

$x-(-i)=x+i\ \text{ and }\ x-i$

are two factors of this function.

So, the function of least degree is

$f(x)=\left(x+\dfrac{1}{3}\right)(x+i)(x-i)=\left(x+\dfrac{1}{3}\right)(x^2-i^2)=\\ \\ =\left(x+\dfrac{1}{3}\right)(x^2+1)=\dfrac{1}{3}(3x+1)(x^2+1)=\dfrac{1}{3}(3x^3+x^2+3x+1)$

If the polynomial function must be with integer coefficients, then it has a form

$f(x)=3x^3+x^2+3x+1$

4 0

4 years ago

Helppppp PLEASEEEEE VERY ARGENTTTT

vekshin1

Answer:

D; x -y = 22

3x + 2y = 246

Step-by-step explanation:

Let us have the bigger number as x and the smaller as y

The difference between the two is 22

Thus, we have it that;

x - y = 22

twice the smaller number; 2(y) added to thrice the larger 3(x) equals 246

3x + 2y = 246

So we have the two equations as;

x -y = 22

3x + 2y = 246

4 0

3 years ago