X2 + X-8 What is the degree

3. Maxine has assets worth $145,000 and liabilities totaling $75,000. What is Maxine's net worth?

BabaBlast [244]

Answer: C.) $70,000

Explanation:
Assets - Liabilities = Net worth
$145,000 - $75,000 = $70,000

5 0

3 years ago

Read 2 more answers

Help me please thx so much have a good one

Mademuasel [1]

Answer: y=1/2x-17

Step-by-step explanation:

8 0

3 years ago

The fish population in a certain lake rises and falls according to the formula F = 3000(23 + 11t − t2). Here F is the number of

Gemiola [76]

Answer:

a) On January 1, 2017 the fish population will be the same as the initial population.

b) On September 18th, 2018 the fish population will be zero.

Step-by-step explanation:

Hi there!

a) First, let´s write the function:

F(t) = 3000(23 + 11t − t²)

The population on January 1, 2006 is the population at t = 0. Then:

F(0) = 3000(23 + 11· 0 - 0²)

F(0) = 3000 · 23 = 69000

This will be the population every time at which t² - 11t = 0. Then let´s find the other value of t (besides t = 0) that makes that expression to be zero:

t² - 11t = 0

t(t - 11) = 0

t = 0

and

t - 11 = 0

t = 11

On January 1, 2017 (2006 + 11), the fish population will be the same as the initial population.

b) We have to obtain the value of t at which F(t) = 0

F(t) = 3000(23 + 11t − t²)

0 = 3000(23 + 11t − t²)

divide both sides of the equation by 3000

0 = 23 + 11t − t²

Let´s solve this quadratic equation using the quadratic formula:

a = -1

b = 11

c = 23

x = [-b ± √(b² - 4ac)] / 2a

x = 12.8 ( the other value of x is negative and therefore discarded).

After 12.8 years all the fish in the lake will have died.

If 1 year is 12 months, 0.8 years will be:

0.8 years · 12 months/year = 9.6 months

If 1 month is 30 days, 0.6 month will be:

0.6 month · 30 days / month = 18 days

All the fish will have died after 12 years, 9 months and 18 days from January 1, 2006. That is, on September 18th, 2018.

7 0

3 years ago

Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loa

Anastasy [175]

Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

$0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0$

The question also has 3 parts given as

Part a: Sketch the deformed shape for α=0.03, β=-0.01 .

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+αx,y+βy)

Point A'(0+(0.03)(0),0+(-0.01)(0))

Point A'(0,0)

Point B(x=1,y=0)

Point B'(x+αx,y+βy)

Point B'(1+(0.03)(1),0+(-0.01)(0))

Point B'(1.03,0)

Point C(x=1,y=1)

Point C'(x+αx,y+βy)

Point C'(1+(0.03)(1),1+(-0.01)(1))

Point C'(1.03,0.99)

Point D(x=0,y=1)

Point D'(x+αx,y+βy)

Point D'(0+(0.03)(0),1+(-0.01)(1))

Point D'(0,0.99)

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

Part b: Calculate the six strain components.

Solution

Normal Strain Components

$\epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\$

Shear Strain Components

$\gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0$

Part c: Find the volume change

$\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\$

Also the change in volume is 0.0197

For the unit cube, the change in terms of strains is given as

$\Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the strain values are small second and higher order values are ignored so

$\Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\$

As the initial volume of cube is unitary so this result can be proved.

5 0

3 years ago

I request assistance on answering this question

Ket [755]

The answer is going to be6/8

6 0

3 years ago