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2 years ago

Emily spent 40% of her check on rent, 12% on groceries, and 16% on untilities. how much did she have left from her $260 check?

Mathematics

1 answer:

Igoryamba2 years ago

4 0

Answer:

83.20

Step-by-step explanation:

If 0.40 + 0.12 + 0.16 = 0.68

then it would be 260x(1-0.68)=83.2

83.2 complete would be 83.20.

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For which value(s) of x will the rational expression below equal zero? Check all that apply.

geniusboy [140]

Answer:

x = 5 and -2

Step-by-step explanation:

Given the function of x to be;

f(x) = (x-5)(x+2)/x+1

The value of x that will make the function zero can be calculated for by equating the given function to zero first to have;

If f(x) = 0, then;

(x-5)(x+2)/x+1 = 0

Cross multiplying;

(x-5)(x+2) = 0

(x-5) = 0 and x+2 = 0

x = 5 and x = -2

The value of x that will make the function zero is therefore 5 and -2

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3 years ago

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The sum of two numbers is 27. One number is 2 times as large as the other. What are the numbers?

mafiozo [28]

X+y=27
x=2y

2y+y=27
y=9

x=27-y
x=27-9=18

7 0

2 years ago

Explain how you can prove that three weeks is less then 24 days

snow_lady [41]

Answer:

7 is the number of how many days are in a week. So 7 x 3 is 21. That is less than 24 days. 24 days is 3 weeks and 3 days.

Step-by-step explanation:

4 0

3 years ago

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What pattern of 7, 15,27

GarryVolchara [31]

Ok well what I did but I’m not sure if it answers your question but I did 27-15=12-7=5 and that what I got maybe it’s 5 it’s just an assumption

7 0

3 years ago

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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