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Crank
3 years ago
15

Selena has $10 in her bank account and adds $10 of her allowance each month. Cooper has $100 in his account and spends $20 per m

onth on playing cards. How many months will it take for Selena and Cooper to have the same amount in their accounts? How much money will they each have at that time?
Solve using tables

Solve using graphs

Solve using substitution

Solve using elimination


I need this problem answered in every way listed, but if you only say one way that would help me so much!


Also please provide a step by step explanation or something like that if you can.


Thank you in advance :)
Mathematics
2 answers:
maxonik [38]3 years ago
7 0

Answer:

They will never have the same amount.

Step-by-step explanation:

Unless you use a number as a point "." then they will never have the same number. Sorry if this doesn't help you

jeka57 [31]3 years ago
3 0
Selina and cooper with both have $40 in their bake account at the same time in month 3 with Selina adding $10 each month and Copper taking $20 away each month.
STEP-BY-STEP
Month 1. Selina- adds $10= $20
Month 1. Cooper- takes $20= $80
Month 2. Selina - adds 10=$30
Month 2. Cooper- takes $20= $60
Month 3. Selina- adds $10= $40
Month 3. Cooper-takes $20= $40
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Consider two vectors A and B A=14i and B= -4i+8j
DaniilM [7]
Assuming \mathbf a,\mathbf b\in\mathbb R^3, you have

\mathbf a\cdot\mathbf b=(14)(-4)+(0)(8)=-56

so

((\mathbf a\cdot\mathbf b)\,\mathbf i)\cdot\mathbf a=(-56\,\mathbf i)\cdot(14\,\mathbf i)=(-56)(14)=-784

Next,

\mathbf a+\mathbf b=(14\,\mathbf i)+(-4\,\mathbf i+8\,\mathbf j)=10\,\mathbf i+8\,\mathbf j

Then

(\mathbf a+\mathbf b)\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\10&8&0\\-4&8&0\end{vmatrix}=112\,\mathbf k

((\mathbf a+\mathbf b)\times\mathbf b)\times\mathbf k=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&112\\0&0&1\end{vmatrix}=\mathbf 0
4 0
3 years ago
Quality control is an important issue at ACME Company which manufacturers light bulbs. In order to conduct testing of the life h
mrs_skeptik [129]

Answer: 3424.28

Step-by-step explanation:

Given data : 378, 361, 350, 375, 200, 391, 375, 368,  321

Number of data values = 9

Mean :\overline{x}=\dfrac{\sum^n_{i=1} x_i}{n}

\\\\=\dfrac{378+361+350+375+200+391+375+368+321}{9}\\\\=\dfrac{3119}{9}\approx346.56

Variance = \dfrac{\sum^n_{i=1} (x_i-\overline{x})^2}{n-1}

\sum^n_{i=1} (x_i-\overline{x})^2 = (31.44)^2+(14.44)^2+(3.44)^2+(28.44)^2+(-146.56)^2+(44.44)^2+(28.44)^2+(21.44)^2+(-25.56)^2\\\\=27394.2224

Variance = \dfrac{27394.2224}{8}=3424.2775\approx3424.28

4 0
3 years ago
Michael is preparing to plant a garden with a rectangular bed in the center and equal square beds on either side. The entire len
Marizza181 [45]

Answer:

what

Step-by-step explanation:

8 0
3 years ago
A number k divided by 6 identify the operation
grigory [225]

Answer:

÷ (could be x if you multiply by 1/6)

Step-by-step explanation:

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6 0
3 years ago
Rachel's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Rachel $4.30 per pound, and typ
Sergio039 [100]

Step 1

Write out the system of equations

let\text{ x represent pound of Type A coffee and y represent Type B coffee pound }\begin{gathered} 4.30x+5.9y=738.80----(1) \\ x+y=148---(2) \end{gathered}

Step 2

Find the value of one pound of type A coffee, x

\begin{gathered} From\text{ equation 2; x+y=148} \\ Therefore \\ y=148-x \end{gathered}\begin{gathered} Substitute\text{ for y in equation 1} \\ 4.30x+5.9(148-x)=738.80 \\ 4.30x+873.2-5.9x=738.80 \\ 873.2-738.80=5.9x-4.3x \\ 134.4=1.6x \\ x=84pounds \end{gathered}

Therefore the answer will be;

\begin{gathered} x+y=148 \\ 84+y=148 \\ y=64\text{ pounds} \\  \end{gathered}

The pounds of type B coffee she used is 64pounds

The pounds of type A coffee she used is 84 pounds

8 0
2 years ago
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