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eduard
3 years ago
10

Michael has $250. He increases his money by 25% and again by 10 %. What was the overall percentage increase in his money ?

Mathematics
1 answer:
valkas [14]3 years ago
8 0

Answer:

His money increased by 6.25, his overall percentage increase is 35

Step-by-step explanation:

250 x 0.25 x 0.10 = 6.25

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Please help me solve 8x+2y=32 and 10x-5y=-20
Tju [1.3M]

Answer:

(2,8)

x = 2

y = 8

Step-by-step explanation:

<u>The steps are shown in the picture</u>

7 0
3 years ago
5 1/2 + 3 4/5 + 1 1/4 =
ratelena [41]

Answer:

10\frac{11}{20}

Step-by-step explanation:

5\frac{1}{2} + 3\frac{4}{5} + 1\frac{1}{4}

<u>1.Find the LCM(Least Common Multiple).</u>

That would be 20.

<u>2.Multiply so all denominators are equal.</u>

5\frac{10}{20} + 3\frac{16}{20} + 1\frac{5}{20}

<u>3. Add</u>

<u></u>9\frac{31}{20}

<u>4. Simplify</u>

<u></u>10\frac{11}{20}

6 0
2 years ago
Pleas help me for 20 points
Montano1993 [528]

Answer:

240

Step-by-step explanation:

1 foot = 12 inches

4/3 feet = 16 inches

area = length × width

= 15 × 16

= 240 square inches

8 0
3 years ago
Which of the following represents a relation that is NOT a function? A. X 7 -5 10 -7 Y 34 32 40 34 B. X -7 -5 -7 2 Y 34 32 40 34
Nostrana [21]
The x-value of -7 shows up more than once in relation ...
  B. X -7 -5 -7 2 Y 34 32 40 34

_____
When there is more than one y-value for an x-value, the relation is NOT a function.
8 0
4 years ago
Solving separable differential equation DY over DX equals xy+3x-y-3/xy-2x+4y-8​
Ivanshal [37]

It looks like the differential equation is

\dfrac{dy}{dx} = \dfrac{xy + 3x - y - 3}{xy - 2x + 4y - 8}

Factorize the right side by grouping.

xy + 3x - y - 3 = x (y + 3) - (y + 3) = (x - 1) (y + 3)

xy - 2x + 4y - 8 = x (y - 2) + 4 (y - 2) = (x + 4) (y - 2)

Now we can separate variables as

\dfrac{dy}{dx} = \dfrac{(x-1)(y+3)}{(x+4)(y-2)} \implies \dfrac{y-2}{y+3} \, dy = \dfrac{x-1}{x+4} \, dx

Integrate both sides.

\displaystyle \int \frac{y-2}{y+3} \, dy = \int \frac{x-1}{x+4} \, dx

\displaystyle \int \left(1 - \frac5{y+3}\right) \, dy = \int \left(1 - \frac5{x + 4}\right) \, dx

\implies \boxed{y - 5 \ln|y + 3| = x - 5 \ln|x + 4| + C}

You could go on to solve for y explicitly as a function of x, but that involves a special function called the "product logarithm" or "Lambert W" function, which is probably beyond your scope.

8 0
2 years ago
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