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Veseljchak [2.6K]
3 years ago
14

A store bought a glass measuring cup at a cost of $1.50 and marked it up 150%. Later on, the store marked it down 80%. What was

the discount price?
Mathematics
1 answer:
malfutka [58]3 years ago
8 0

Answer:

45 cents

Step-by-step explanation:

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If 40% of x is 8, what is x% of 40?<br> (A) 80<br> (B) 30<br> (C) 10<br> (D) 8
quester [9]

Answer:

d

Step-by-step explanation:

40% = 40/100

40/100 X = 8

X = 800/40 = 20

X% of 40 = 20/100*40 = 8

7 0
4 years ago
Read 2 more answers
Use the commutative property of multiplication to identify which expression is equivalent to 6(2)(x).
11Alexandr11 [23.1K]
The commutative property is when you change the order but not the answer. To better understand what they are asking you, let's let x=1- 6 * 2 * 1=12; Answer A wouldn't be right because there are no 6,2, or x's. Remember you just change the ORDER not the GROUPING so Answer B is wrong too. Now Answer C and Answer D are left. Now you just need to look at the signs. 6(2) just means 6 TIMES 2 so pick the option with multiplication. So your answer is C. Hope I helped!
5 0
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

x>0,y>1 \implies 2|x| + 3|y-1| = 2x + 3(y-1)=6 \implies 2x + 3y = 9

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}.

5 0
2 years ago
What is the solution to the inequality x^2 &gt;64
Gnom [1K]

Answer:

see explanation

Step-by-step explanation:

Consider

x² = 64 ( take the square root of both sides )

x = 8 , x = - 8 ← are the critical points

Then for x² > 64

x < - 8 or x > 8

4 0
3 years ago
I need help with this whole pg please help me
tekilochka [14]
Message me and i’ll help you
7 0
3 years ago
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