To find the answer of this problem, we need to distibute and multiply all of the numbers with each other.

2x * x + 2x * 1 + 9 * x + 9 * 1
= 2x^2 + 2x + 9x + 9
= 2x^2 + 11x + 9

The answer would be the first option.

Hope this helps!

6 0

3 years ago

You have $7.45. You buy 8 items that cost $0.83 each. How much money do you have left

Agata [3.3K]

Answer:

$0.81

Step-by-step explanation:

left

$7.45-8(0.83)=7.45-6.64=0.81$

I hope I have helped you

6 0

3 years ago

Find the vectors T, N, and B at the given point. r(t) = < t^2, 2/3t^3, t >, (1, 2/3 ,1)

maxonik [38]

Answer with Step-by-step explanation:

We are given that

$r(t)=< t^2,\frac{2}{3}t^3,t >$

We have to find T,N and B at the given point t > (1, $2/3$ ,1)

$r'(t)=$

$\mid r'(t) \mid=\sqrt{(2t)^2+(2t^2)^2+1}=\sqrt{(2t^2+1)^2}=2t^2+1$

$T(t)=\frac{r'(t)}{\mid r'(t)\mid}=\frac{}{2t^2+1}$

Now, substitute t=1

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

$T'(t)=\frac{-4t}{(2t^2+1)^2} +\frac{1}{2t^2+1}$

$T'(1)=-\frac{4}{9}+\frac{1}{3}$

$T'(1)=\frac{1}{9}=$

$\mid T'(1)\mid=\sqrt{(\frac{-2}{9})^2+(\frac{4}{9})^2+(\frac{-4}{9})^2}=\sqrt{\frac{36}{81}}=\frac{2}{3}$

$N(1)=\frac{T'(1)}{\mid T'(1)\mid}$

$N(1)=\frac{}{\frac{2}{3}}=$

$N(1)=$

$B(1)=T(1)\times N(1)$

$B(1)=\begin{vmatrix}i&j&k\\\frac{2}{3}&\frac{2}{3}&\frac{1}{3}\\\frac{-1}{3}&\frac{2}{3}&\frac{-2}{3}\end{vmatrix}$

$B(1)=i(\frac{-4}{9}-\frac{2}{9})-j(\frac{-4}{9}+\frac{1}{3})+k(\frac{4}{9}+\frac{2}{9})$

$B(1)=-\frac{2}{3}i+\frac{1}{3}j+\frac{2}{3}k$

$B(1)=\frac{1}{3}$

5 0

3 years ago

Two quadratic functions are shown.

slava [35]

Answer:

Option (1) is the correct option.

Step-by-step explanation:

Function 1,

f(x) = 4x² + 8x + 1

= 4(x² + 2x) + 1

= 4(x² + 2x + 1 - 1) + 1

= 4(x² + 2x + 1) - 4 + 1

f(x) = 4(x + 1)² - 3

This graph opens up with the vertex or minimum point at (-1, -3)

So, the minimum value of the function is (-3) at x = -1.

Function (2)

From the given table minimum value of the function is 0 at x = -1 or minimum point as (-1, 0)

Therefore, Function 1 has the least minimum value and its coordinates are (-1, -3)

Option (1) is the correct option.

7 0

3 years ago