All categories

2 years ago

I'm trying to figure this out, but I can't get it

Mathematics

2 answers:

oksano4ka [1.4K]2 years ago

8 0

$\\ \sf\longmapsto \sqrt{16r^6}$

$\\ \sf\longmapsto \sqrt{16}\sqrt{r^6}$

$\\ \sf\longmapsto 4r^{6\times \dfrac{1}{2}}$

$\\ \sf\longmapsto 4r^3$

Option B is correct

Katen [24]2 years ago

5 0

Answer:

$\sqrt{16 {r}^{6} } \\ ( \sqrt{16 {r}^{6} } )^{2} \\ (4 {r}^{3} )$

You might be interested in

Find the angle θθ between the vectors v=2i+kv=2i+k, w=j−3kw=j−3k.

kaheart [24]

Hello :
v . w = | | v | | × | | w | | cos(θ) ....(1)
v(2,0,1) w(0,1,-3)

v . w = (2)(0)+(0)(1)+(1)(-3) = - 3
| | v | | = √((2)²+(0)²+(1)²) = √5
| | w | = √((0)²+(1)²+(-3)²) = √10
by (1) :
cos(θ) = (v . w ) / | | v | | × | | w | |

cos(θ) = = - 3/√50
θ =..... calculate by 2ind function ( calculator)

4 0

3 years ago

Michael can buy 4 sandwiches and 3 cups of coffee for $23.75. Amy can buy 2 sandwhiches and 5 cups of coffee for $19.75 how much

Kobotan [32]

Answer: The cost of one sandwich is $4.25 and the cost of one cup of coffee is $2.25.

Step-by-step explanation:

Let x be the cost of one sandwich and y be the cost of one cup of coffee.

By considering the given situation, we have the following equations :-

$4x+3y=23.75.................(1)\\\\2x+5y=19.75................(2)$

Multiply 2 on both sides of equation (2) , we get

$4x+10y=39.5................(3)$

Subtract equation (1) from equation (3), we get

$7y=15.75\\\\\Rightarrow\ y=\dfrac{15.75}{7}=2.25$

Put value of y in (1), we get

$4x+3(2.25)=23.75\\\\\Rightarrow\ 4x+6.75=23.75\\\\\Rightarrow\ 4x =17\\\\\Rightarrow\ x=4.25$

Hence, the cost of one sandwich is $4.25 and the cost of one cup of coffee is $2.25.

8 0

2 years ago

PLEASE HELP (30 POINTS) Solve the rational equation x/3 = x^2/x + 5 , and check for extraneous solutions.

anygoal [31]

Option C: $x=0$ and $x=\frac{5}{2}$ are the solutions.

Explanation:

The equation is $\frac{x}{3} =\frac{x^{2} }{x+5}$

We shall determine the value of x, by simplifying the equation.

$\begin{aligned} x(x+5) &=3 x^{2} \\ x^{2}+5 x &=3 x^{2} \\ 2 x^{2}-5 x &=0 \\ x(2 x-5) &=0 \end{aligned}$

Thus, $x=0$ and $x=\frac{5}{2}$ are the solutions.

Now, let us check whether the solutions are extraneous solutions.

Let us substitute $x=0$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}&\frac{0}{3}=\frac{0^{2}}{0+5}\\&0=\frac{0}{5}\\&0=0\end{aligned}$

Thus, both sides of the equation are equal.

Hence $x=0$ is a true solution.

Now, Let us substitute $x=\frac{5}{2}$ in the original equation to check whether both sides of the equation are equal.

$\begin{aligned}\frac{\left(\frac{5}{2}\right)}{3} &=\frac{\left(\frac{5}{2}\right)^{2}}{\left(\frac{5}{2}\right)+5} \\\frac{5}{6} &=\frac{\left(\frac{25}{4}\right)}{\left(\frac{15}{2}\right)} \\\frac{5}{6} &=\frac{5}{6}\end{aligned}$