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goldenfox [79]
3 years ago
8

Turn 276% into a mixed number

Mathematics
1 answer:
AlladinOne [14]3 years ago
6 0
Mixed number is (wholenumber)+(properfraction)
a proper fraction is a/b where a<b and b≠0

so

percent means parts out of 100
x%=x/100

so
276%=
276/100=
200/100+76/100=
2+76/100=
2+38/50=
2+19/25

2 and 19/25
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Answer:

The answer is 8 are Black and 4 are Latino

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Sergeu [11.5K]

Answer: the tax is $2.16 so total would be $74.16 if the original ticket price was $72.00

Step-by-step explanation:

T + 0.03

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72 • 0.03 = 2.16

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The equation 2(4x + 1) = 4(2x + 5) has no solutions. Which of the following best explains why?
Vladimir79 [104]

Answer:

See below

Step-by-step explanation:

2(4x + 1) = 4(2x + 5)

8x + 2 =8x + 20

\nexists x \in \mathbb{R} \text{ That makes the equality true}

The equation has no solution because no matter what the value we plug for x in the equation, it will not be true and will lead to a contradiction. This is the meaning of not having a solution.

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4 years ago
Find the area of the region defined by the region defined by the inequality 2|x| + 3|y-1| ≤ 6
EleoNora [17]

If x and y-1 have the same sign, then either

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or

x

If x and y-1 have opposite sign, then

x>0,y

or

x1 \implies 2|x| + 3|y-1| = -2x + 3(y-1) = 6 \implies 2x-3y = -9

This is to say that the region has boundaries given by these two sets of parallel lines, so we can equivalently describe the region with the set

R = \left\{(x,y) \mid -3\le2x+3y\le9 \text{ and } -9\le2x-3y\le3\right\}

The area of R is given by the double integral

\displaystyle \iint_R dx\,dy

To compute the area, change the variables to

\begin{cases}u = 2x + 3y \\ v = 2x - 3y\end{cases} \implies \begin{cases}x = \frac14(u+v) \\ y = \frac16(u-v)\end{cases}

The Jacobian for this transformation is

J = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} = \begin{bmatrix}1/4 & 1/4 \\ 1/6 & -1/6\end{bmatrix}

with determinant \det(J) = -\frac1{12}. Then the integral transforms to

\displaystyle \iint_R dx\,dy = \iint_R |J| \, du \, dv = \frac1{12} \int_{-3}^9 \int_{-9}^3 dv\, du

which is 1/12 the area of a square with side length 12. Hence the integral evaluates to

\displaystyle \iint_R dx\,dy = \frac1{12}\times12^2 = \boxed{12}.

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