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Ostrovityanka [42]

2 years ago

5

Determine if the expression -a^3b^5 is a polynomial or not. If it is a polynomial, state

1 answer:

Dmitry_Shevchenko [17]2 years ago

5 0

I think so and degree of 5

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Help plz!!!!!?!!!?!??!????

aleksandrvk [35]

700ml=23.6698fl
64-23.6698=40.3302
40.3302fl

5 0

2 years ago

10. How many three-topping pizzas can be ordered from the list of toppings below? Did you calculate the number of permutations o

Nimfa-mama [501]

Step-by-step explanation:

Total numbers of pizza toppings = 12

Number of three topping pizzas can be ordered are:

$P^{n}_{k}=\frac{n!}{(n-k)!}$

where = n = number of elements

k = number of elements choosen

n= 12 , k = 3

$\frac{12!}{(12-3)!} =\frac{12\times 11\times 10\times 9!}{9!}= 1,320$

We calculated the number of permutations. We made this choice because in permutation order of element matters but in combination its not.

8 0

2 years ago

24% of what is 6?

IrinaK [193]

Answer:

25

Step-by-step explanation:

Percentage Calculator: 6 is what percent of 24? = 25.

7 0

2 years ago

Read 2 more answers

How to solve this problem??

slega [8]

$\frac{2}{3} = \frac{18}{x+5}
$
multiply each side by $x+5$
$\frac{2x+10}{3}=18$
multiply each side by 3

8 0

2 years ago

Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)

vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

$x^{\frac{2}{3}}+y^{\frac{2}{3}}=4$

We have to find the equation of tangent line to the given curve at point $(-3\sqrt3,1)$

By using implicit differentiation, differentiate w.r.t x

$\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0$

Using formula : $\frac{dx^n}{dx}=nx^{n-1}$

$\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}$

$\frac{dy}{dx}=-\frac{x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}}$

Substitute the value x= $-3\sqrt3,y=1$

Then, we get

$\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}$

$\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}$

Slope of tangent=m= $\frac{1}{\sqrt3}$

Equation of tangent line with slope m and passing through the point $(x_1,y_1)$ is given by

$y-y_1=m(x-x_1)$

Substitute the values then we get

The equation of tangent line is given by

$y-1=\frac{1}{\sqrt3}(x+3\sqrt3)$

$y-1=\frac{x}{\sqrt3}+3$

$y=\frac{x}{\sqrt3}+3+1$

$y=\frac{x}{\sqrt3}+4$

This is required equation of tangent line to the given curve at given point.

8 0

3 years ago