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SIZIF [17.4K]
4 years ago
9

If your expenses are greater than your income you need to a. re-do your budget. c. save more money. b. reduce your spending. d.

all of the above.
Mathematics
2 answers:
qaws [65]4 years ago
7 0
B)reduce your spending/c) all of the above
Novay_Z [31]4 years ago
3 0

Answer:

d. all of the above

Step-by-step explanation:

When your expenses are higher than your income, this creates a lot of problems like - if this thing happens without any check, then soon your savings are gone and you are in debt.

In such situation, you will have to earn and save more money and you will need to cut back your expenses and scrutinize and re do your budget.

So, basically its all the options given here.

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 2) The second step is to complete the squares.

 3) The third step is factor the quadratic equation

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the first step in finding the center of the circle by completing the square is:
 
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3 years ago
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3 0
3 years ago
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PLEASE SOMEONE HELP ILL GIVE BRAINLIEST!
grin007 [14]

Answer:

Part A

The rotation of the point (x, y) 180° about the origin gives the point (-x, -y)

Therefore, we have the points, ΔLMN, L(1, 1), and M(2, 2), and N(3, 3)

The coordinates of the vertices of the image, ΔL'M'N' are L'(-1, -1), and M'(-2, -2), and N'(-3, -3)

Please find attached the graph of triangle ΔLMN and ΔL'M'N'

Part B

The lines drawn through L and L' and through M and M' are colinear

Part C

A line is defined by two points, such as <em>L </em>and <em>M.</em> Rotation of a line through 180° about the origin will give an image location on the same path as the extension of the original line.

The third point, <em>N</em>, however, defines a plane, and the rotation of a plane by 180° will give an image which is turned upside down with regards to the preimage

Therefore, a trough in the preimage becomes a peak in the image and the lines drawn through N and N' crosses the colinear lines drawn throgh M and M' and L and L'

Step-by-step explanation:

5 0
3 years ago
(1 point) a tank contains 1060 l of pure water. a solution that contains 0.06 kg of sugar per liter enters the tank at the rate
kirill115 [55]
(a) There is 0 kg of sugar in the tank at the beginning since it contains pure water at the start. The sugar only comes from the solution.

(b)

S' = f(t,S) = \left(0.06 \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) - \left(\dfrac{S}{1060} \dfrac{\text{kg}}{\text{L}}\right)\left(9\dfrac{\text{L}}{\text{min}}\right) \ \Rightarrow \\ \\ S' = 0.54 \text{ kg}/\text{min} - \dfrac{9S}{1060}

So yes, you enter S' = 0.54 - (9S/1060)

(c)

\displaystyle\frac{dS}{dt} = 0.54 - \frac{9S}{1060} \ \Rightarrow\ \frac{dS}{dt} = \frac{572.4 - 9S}{1060}\ \Rightarrow\ \dfrac{dS}{572.4 - 9S} = \frac{1}{1060} dt\ \Rightarrow \\ \\&#10;\int \dfrac{dS}{572.4 - 9S} = \int \frac{1}{1060} dt\ \Rightarrow\textstyle\ -\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t + C \\ \\&#10;S(0) = 0 \ \Rightarrow\ -\frac{1}{9}\ln|572.4 - 0| = \frac{1}{1060}(0) + C\  \Rightarrow\ C = -\frac{1}{9} \ln 572.4

-\frac{1}{9}\ln|572.4 - 9S| = \frac{1}{1060}t  -\frac{1}{9} \ln 572.4\ \Rightarrow \\ \\&#10;\ln|572.4 - 9S| = \ln 572.4 - \frac{9}{1060}t \ \Rightarrow \\ \\&#10;|572.4 - 9S| = e^{\ln 572.4 - 9t/1060}\ \Rightarrow \\ \\&#10;572.4 - 9S= \pm 572.4 e^{-9t/1060}\ \Rightarrow \\ \\&#10;S = \frac{-1}{9}\left(-572.4 \pm 572.4 e^{-9t/1060}\right)

But only (+) satisfies S(0) = 0

S= -\frac{1}{9}\left(-572.4 + 572.4 e^{-9t/1060}\right) \\ \\&#10;S= 63.6 - 63.6 e^{-9t/1060}\text{ kg}

Enter
in S = 63.6 - 63.6 * e^(-9t/1060)

3 0
3 years ago
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