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3 years ago

Please help me answer this ahh

Mathematics

1 answer:

Katarina [22]3 years ago

4 0

Your numerator is correct, but not your denominator.

-(1/4)(log to the base 3) of z is equivalent to z raised to the power -(1/4). Thus, you must divide your numerator (given above) by z^(1/4).

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Alan bought two bikes. He sold one to Beth for $300 taking a 25% loss. He also sold one to Greta for $300 making a 25% profit. D

Mashutka [201]

No Alan did not break even

Alan incured a loss of 6.25 %

<em><u>Solution:</u></em>

Given that,

Alan bought two bikes

He sold one to Beth for $300 taking a 25% loss

He also sold one to Greta for $300 making a 25% profit

When a person sells two similar items, one at a gain of say x%, and the other at a loss of x%, then the seller always incurs a loss

Which is given as:

$loss\ \% = (\frac{x}{10})^2$

Here, x = 25

$loss\ \% = (\frac{25}{10})^2\\\\loss\ \% = 2.5^2\\\\loss\ \% = 6.25$

Thus the loss percentage is 6.25 %

8 0

3 years ago

AB=2y+1, BC=y+1,CD=7x-3,DA=3x what is x and y

sattari [20]

I attached the picture associated with this question.

Answer:
x = 2
y = 5

Explanation:
ABCD is a parallelogram. This means that each two opposite sides are equal.
This means that:
1- AB = CD
2y + 1 = 7x - 3 ...........> equation I
2- AD = BC
3x = y + 1
This can be rewritten as:
y = 3x - 1............> equation II

Substitute with equation II in equation I and solve for x as follows:
2y + 1 = 7x - 3 ...........> equation I
2(3x - 1) + 1 = 7x - 3
6x - 2 + 1 = 7x - 3
6x - 1 = 7x - 3
7x - 6x = -1 + 3
x = 2

Substitute with x in equation I to get y as follows:
y = 3x - 1
y = 3(2) - 1
y = 6 - 1
y = 5

Hope this helps :)

3 0

3 years ago

What is the equation of the line that passes through the point (-2,3) and that is perpendicular to the line represented by 3x-2y

motikmotik

Answer:

3y + 2x = 5

Step-by-step explanation:

3x-2y = -2

2y = 3x + 2

y = (3/2)x + 1

Slope is 3/2

Slope of the Perpendicular line is -2/3

y = (-2/3)x + c

x = -2 , y = 3

3 = (-2/3)(-2) + c

3 = 4/3 + c

c = 3 - 4/3 = 5/3

y = (-2/3)x + 5/3

3y = -2x + 5

3y + 2x = 5

5 0

3 years ago

Which graph correctly solves the system of equations below? y = −x2 + x + 7 y = x2 + 3x + 7

antiseptic1488 [7]

Answer:

(-1.703, 2.396)

(1.37, 6.493)

Step-by-step explanation:

We can easily solve this question by using a plotting tool or any graphing calculator,

The solution to the system of equation is given by the intersection points between the graphs of

y1 = −x^2 + x + 7

y2 = x^2 + 3x + 7

Please, see attached picture below.

The solutions are:

(-1.703, 2.396)

(1.37, 6.493)

5 0

3 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