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3 years ago

Please help ASAP ASAP please please

Mathematics

1 answer:

Doss [256]3 years ago

8 0

Answer:

25,65

Step-by-step explanation:

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A table and 8 chairs weigh 235.68 lb together. If the table weighs 157.84 lb, what is the weight of one chair in pounds?

Alona [7]

Answer: 9.73 lb each chair

Step-by-step explanation:

You substract the weight of the table alone from the full weight:

235.68 - 157.84=77.84

Then you divide 77.84 by 8 chairs and you obtain the weight of each individual chair

77.84/8= 9.73 the weight of each individual chair

7 0

3 years ago

Read 2 more answers

What is the area of a rectangle with vertices (Negative 8, Negative 2), (Negative 3, Negative 2), (Negative 3, Negative 6), and

kap26 [50]

Answer:

20 unit square is the area of a rectangle with given vertices.

Step-by-step explanation:

Area of the rectangle with vertices : Say ABCD:

$A(-8,-2), B(-3,-2), C (-3,-6), D(-8,-6)$

Distance formula : $(x_1,y_1) (x_2,y_2)$

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Length of the rectangle= AB

$AB=\sqrt{(-3-(-8))^2+(-2-(-2))^2}=5 units$

Breadth of the rectangle= BC

$BC=\sqrt{(-3-(-3))^2+(-6-(-2))^2}=4 units$

Area of rectangle = Length × Breadth = AB × BC

$A= 5 units \times 4 units = 20 unit^2$

20 unit square is the area of a rectangle with given vertices.

5 0

3 years ago

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Stan's savings account has a balance of $4706. After 4 years, what will the amount of interest be at 4% compounded quarterly? I

antiseptic1488 [7]

Answer: $812.16

Step-by-step explanation:

The concept of compound interest is that interest is added back to the principal sum so that interest is gained on that already-accumulated interest during the next compounding period.

Interest can be compounded on any given frequency schedule, from continuous to daily to annually. When incorporating multiple compounds per period (monthly compounding or quarterly compounding, etc), the general formula looks like this:

$\boxed{A = P(1 +\frac{r}{n})^{n*t}}$

A = the future value of the investment/loan, including interest

P = the principal investment amount (the initial deposit or loan amount)

r = the annual interest rate (decimal)

n = number of times interest is compounded per unit "t"

t = the time the money is invested or borrowed for

<h3>(Using the data provided in the question )</h3>

P = $4706

r = 4/100 = 0.04

n = 4 (compounded quarterly)

t = 4 years 

$A = 4706(1 +\frac{0.04}{4})^{4*4}$

A = $4706(1 + 0.04/4)¹⁶

A = $4706(1 + 0.01)¹⁶

A = $4706(1.01)¹⁶

A = $4706(1.1725787) = $5518.16

This formula gives the combined principal amount and its compound interest, so subtract the principal amount to get just the compound interest.

$5518.16 - $4706 = $812.16

Answer: $812.16

$\textit{\textbf{Spymore}}$

6 0

3 years ago

Exercise 3.3.2: Verify that the system x® 0 1 3 3 1 x® has the two solutions 1 1 e 4t and 1 −1 e −2t a) . b) Write down the gene

nikklg [1K]

Answer:

The general solution of the system is

$X(t)=A\left(\begin{array}{c}1 &-1 \end{array}\right)e^{-2t} + B\left(\begin{array}{c}1 &1 \end{array}\right)e^{4t}\\$

Step-by-step explanation:

To verify that the system $X^{'} =\left[\begin{array}{cc}1&3\\3&1\end{array}\right] X$ has the solutions given,

First, we determine the Eigen Values β of the matrix of the form $X^{'} =A X$ Using |A-βI|=0, where I is the Identity Matrix, we have:

$|A-\beta I|=0$

$|\left(\begin{array}{cc}1&3\\3&1\end{array}\right )-\left(\begin{array}{cc}\beta &0\\0&\beta \end{array}\right )|=0$

Subtracting matrices

$\left|\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right |=0$

Taking the determinant

$(1-\beta)(1-\beta)-(3X3)=0\\1-\beta-\beta+\beta^{2}-9=0\\\beta^{2}-2\beta-8=0\\\beta^{2}-4\beta+2\beta-8=0\\\beta(\beta-4)+2(\beta-4)=0\\(\beta-4)((\beta+2)=0\\\beta=4 or -2$

We determine the eigen vector using $\left(\begin{array}{cc}1-\beta &3\\3&1-\beta \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0$

When $\beta$ = -2,

$\left(\begin{array}{cc}3 &3\\3&3 \end{array}\right)\left(\begin{array}{c}c_{11} &c_{12} \end{array}\right)=0$ which implies that $3c_{11}+3c_{12}=0$ and thus

$If c_{11}=1, c_{12}=-1$

When $\beta$ = 4,

$\left(\begin{array}{cc}-3 &3\\3&-3 \end{array}\right)\left(\begin{array}{c}c_{21} &c_{22} \end{array}\right)=0$ which implies that $3c_{21}-3c_{22}=0$ and thus