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3 years ago

A student estimates that it takes 2 hours to read 25 pages. At this rate, how many hours will she need to read 425 pages?

Mathematics

2 answers:

wolverine [178]3 years ago

7 0

Answer:

34 hours

Step-by-step explanation:

2:25 = x:425

25/2 = 12.5

425/12.5 = 34

so it is 34:425

She will need to read for 34 hours to read 425 pages

Aliun [14]3 years ago

5 0

Answer:

25/425 = 2/x

Step-by-step explanation:

Cross multiply and simplify

850=25x

34=x

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(a) (4,4) is a point on a locus whose equation is y2 = kx. Show that (16,8) is another point on the locus.

Vlad [161]

Step-by-step explanation:

plug the coordinate points into the equations and solve for the remaining variable.

b) (2,3) - 5(2) + b(3) = 16

10 + 3b = 16. 3b= 6. b = 2

(0,8) - 5(0) + 2(8) = 16. 0+ 16 = 16. ☆

a) (4,4) assume based on results you meant y squared. (y^2)

(4)^2 = 4k. 16 = 4k. k = 4

(8)^2 = 4(16). 64 = 64. ☆

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3 years ago

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Evaluate.

tekilochka [14]

Step-by-step explanation:

W=-6, x=1.2, and z=-6/7

(W²x-3)÷10-z

we substitute

((-6²)(1.2)-3)÷10-(-6/7)

((-36)(1.2)-3) ÷10-(-6/7)

(-43.2-3) ÷10(6/7)

(-46.2)÷60/7

-46.2÷60/7

-46.2*7/60

-46.2/1*7/60

-323.4/60

-5.39

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3 years ago

Lifetime Fitness charges $90 a month for membership plus $50 per session with a trainer. Which equation can be used to determine

Taya2010 [7]

Answer:

50x / 30/x i think

Step-by-step explanation:

7 0

3 years ago

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Please help me with this.<br>Thank you

Solnce55 [7]

Answer:

(17+4)/21. as denominators are same

21/21

3 0

2 years ago

Find an equation of the tangent plane to the given parametric surface at the specified point.

Neko [114]

Answer:

Equation of tangent plane to given parametric equation is:

$\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

Step-by-step explanation:

Given equation

$r(u, v)=u cos (v)\hat{i}+u sin (v)\hat{j}+v\hat{k}$ ---(1)

Normal vector tangent to plane is:

$\hat{n} = \hat{r_{u}} \times \hat{r_{v}}\\r_{u}=\frac{\partial r}{\partial u}\\r_{v}=\frac{\partial r}{\partial v}$

$\frac{\partial r}{\partial u} =cos(v)\hat{i}+sin(v)\hat{j}\\\frac{\partial r}{\partial v}=-usin(v)\hat{i}+u cos(v)\hat{j}+\hat{k}$

Normal vector tangent to plane is given by:

$r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]$

Expanding with first row

$\hat{n} = \hat{i} \begin{vmatrix} sin(v)&0\\ucos(v) &1\end{vmatrix}- \hat{j} \begin{vmatrix} cos(v)&0\\-usin(v) &1\end{vmatrix}+\hat{k} \begin{vmatrix} cos(v)&sin(v)\\-usin(v) &ucos(v)\end{vmatrix}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u(cos^{2}v+sin^{2}v)\hat{k}\\\hat{n}=sin(v)\hat{i}-cos(v)\hat{j}+u\hat{k}\\$

at u=5, v =π/3

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

at u=5, v =π/3 (1) becomes,

$r(5, \frac{\pi}{3})=5 cos (\frac{\pi}{3})\hat{i}+5sin (\frac{\pi}{3})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=5(\frac{1}{2})\hat{i}+5 (\frac{\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

$r(5, \frac{\pi}{3})=\frac{5}{2}\hat{i}+(\frac{5\sqrt{3}}{2})\hat{j}+\frac{\pi}{3}\hat{k}$

From above eq coordinates of r₀ can be found as:

$r_{o}=(\frac{5}{2},\frac{5\sqrt{3}}{2},\frac{\pi}{3})$

From (2) coordinates of normal vector can be found as

$n=(\frac{\sqrt{3} }{2},-\frac{1}{2},1)$

Equation of tangent line can be found as:

$(\hat{r}-\hat{r_{o}}).\hat{n}=0\\((x-\frac{5}{2})\hat{i}+(y-\frac{5\sqrt{3}}{2})\hat{j}+(z-\frac{\pi}{3})\hat{k})(\frac{\sqrt{3} }{2}\hat{i}-\frac{1}{2}\hat{j}+\hat{k})=0\\\frac{\sqrt{3}}{2}x-\frac{5\sqrt{3}}{4}-\frac{1}{2}y+\frac{5\sqrt{3}}{4}+z-\frac{\pi}{3}=0\\\frac{\sqrt{3}}{2}x-\frac{1}{2}y+z=\frac{\pi}{3}$

5 0

3 years ago