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OLEGan [10]
2 years ago
12

A display case of hat pins are marked 5 for $3. If Patty has $21, how many hat pins can Patty get? (assume no other taxes and fe

es)
Mathematics
1 answer:
Nonamiya [84]2 years ago
4 0
Answer: 35

Explanation: If patty has $21, and $3 is the price for 5 hat pins, then: 21/3=7, so she can buy 7 packages which cost $3 and contain 5 pins, and because the question asked the number of pins, we will multiply 7 • 5 and that equals 35
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Which transformations could have been used to create A'B'C'? Choose all that apply.
Eduardwww [97]

The correct transformation is a rotation of 180° around the origin followed by a translation of 3 units up and 1 unit to the left.

<h3>Which transformation is used to get A'B'C'?</h3>

To analyze this we can only follow one of the vertices of the triangle.

Let's follow A.

A starts at (3, 4). If we apply a rotation of 180° about the origin, we end up in the third quadrant in the coordinates:

(-3, -4)

Now if you look at A', you can see that the coordinates are:

A' = (-4, -1)

To go from (-3, -4) to (-4, -1), we move one unit to the left and 3 units up.

Then the complete transformation is:

A rotation of 180° around the origin, followed by a translation of 3 units up and 1 unit to the left.

If you want to learn more about transformations:

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6 0
1 year ago
A survey was administered to high school seniors in Anytown. According to the survey results, fewer than 0.5% of the students dr
Feliz [49]

Answer: high test-retest reliability

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8 0
2 years ago
The universal set in this diagram is the set of integers from 1 to 15. place the integers in the correct place in the venn diagr
neonofarm [45]

Answer:


Step-by-step explanation:

Given :  universal set in this diagram is the set of integers from 1 to 15.

Solution :

The intersection of odd integer,multiples of 3 and Factors of 15 are 3,15

The intersection of odd integer and Factors of 15 are 1,5

The intersection of odd integer,multiples of 3 is 9

The remaining multiples of 3 are 6,12

The remaining odd integers are 7,11,13

Now the remaining integers are 2,4,8,10,14 and these integers must be placed in the boxes outside the circles Since they does not belong any intersection or odd integer or factor of 15 .

Refer the attached figure for the answer.

3 0
2 years ago
Read 2 more answers
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Anna71 [15]

Answer:

D. (2x)(3x) + 5(3x - 2)

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5 0
3 years ago
Is h(x)=3x^3+2x^2+5 even, odd, or neither?
pochemuha

Answer:

Neither even or odd

Step-by-step explanation:

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2 years ago
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