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4 years ago

Four pounds of apples cost 3.88. what is the price per pound

Mathematics

1 answer:

Novay_Z [31]4 years ago

3 0

Answer:

0.97 cents

Step-by-step explanation:

4/3.88=0.97

price per pound=0.97 cents

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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

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

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

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

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

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

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

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

$N(1)=$

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

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

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

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

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

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3 years ago

The graph below shows a line segment XY: graph of line segment XY with endpoints at negative 1 comma 3 and 1 comma negative 1 Wh

yan [13]

Slope = (y2 - y1) / (x2 - x1)
(-1,3)....x1 = -1 and y1 = 3
(1,-1) ...x2 = 1 and y2 = -1
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3 years ago

Read 2 more answers

Find all angles, 0≤θ<360, that satisfy the equation below, to the nearest 10th of a degree.

natta225 [31]

Since cosine is negative in the second and 3rd quadrant, the required angles are 120 and 240 degrees

<h3>Trigonometry identity</h3>

Trigonometry identities are expressed as a function of cosine, sine and tangent.

Given the trigonometry expression shown

4cos2θ+9=−14cosθ

Equate to zero

4cos2θ+9 + 14cosθ = 0

According to trig identity

cos2θ = 2cos²θ - 1

Substitute to have:

4(2cos²θ - 1)+9 + 14cosθ = 0

Expand

8cos²θ - 4 + 9 + 14cosθ = 0

8cos²θ+ 14cosθ + 5 = 0

let P = cosθ to have;

8P² + 14P + 5 = 0

Factorize the result

8P² + 10P + 4P + 5 = 0

2P(4P+5)+1(4P+5)=0
(2P+1) = 0 and 4P+5 = 0
2P = -1 and P = -5/4

P = -1/2 and -5/4

Recall that P = cosθ

If P = -1/2

cosθ = -1/2

θ = -60

Since cosine is negative in the second and 3rd quadrant, the required angles are 120 and 240 degrees

Learn more on trigonometry identity here: brainly.com/question/24349828

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2 years ago