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Alenkasestr [34]
3 years ago
6

What is an dividend in a math equation?

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
4 0
Example:

35 ÷ <u>5</u> = <em>7</em>

35 = dividend
<u>5 = divisor</u>
<em>7 = quotient </em>
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Which word best describes the degree of overlap between the two data sets?
snow_tiger [21]

Answer:

Moderate

Step-by-step explanation:

I think sorry if its wrong

6 0
3 years ago
What is A decreased by 20% and then increased by 20% of that? Please show your work.
umka21 [38]

Answer: \\ A - 20\%A = A - 0.2A = 0.8A \\ 0.8A + 20\%0.8A = 0.8A + 0.2 \times 0.8A = 0.8 + 0.16A = 0.96A

8 0
3 years ago
A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be
irga5000 [103]

Answer:

The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

Step-by-step explanation:

Let C = (h,k) the coordinates of the center of the circle, which must be transformed into C'=(h', k') by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

(x-h)^{2}+(y-k)^{2} = r^{2}

Where r is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):

C''(x,y) = C(x, y) + U(x,y) (Eq. 1)

Where U(x,y) is the translation vector, dimensionless.

If we know that C(x, y) = (h,k) and U(x,y) = (4, 0), then:

C''(x,y) = (h,k)+(4,0)

C''(x,y) =(h+4,k)

2) <em>Reflection over the x-axis</em>:

C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)] (Eq. 2)

Where O(x,y) is the reflection point, dimensionless.

If we know that O(x,y) = (h+4,0) and C''(x,y) =(h+4,k), the new point is:

C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]

C'(x,y) = (h+4, 0)-(0,k)

C'(x,y) = (h+4, -k)

And thus, h' = h+4 and k' = -k. The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

<em />

4 0
3 years ago
Can anybody please help solving this problem?
pantera1 [17]
Area of the shaded region = Area <span>of a square - Area of a circle
</span>
Side of a square = 2r
Area of a square = (2r)² = 4r²

Area of a circle = πr²

Area of the shaded region = 4r² - πr² = <span>r²(4-π)</span>
7 0
3 years ago
***URGENT*** PLEASE HELP ME ASAP, ITS DUE TODAY!!!
vladimir2022 [97]

Answer:

Construct MN.

Since M is the midpoint of OA, OM = MA

Similarly, N is the midpoint of OB.

Thus, ON = NB.

Now, in Δs OMN and OAB,

∠MON = ∠AOB (common angle)

(sides are in proportional ratio; OA = 2OM and OB = 2ON)

∴ Δs OMN and OAB are similar (2 sides are in proportion, with the included angle)

Since they are similar, then ∠OMN = ∠OAB (corresponding angles of similar triangles are equal)

But since ∠OMN = ∠OAB, then that means MN || AB (corresponding angles of two lines must be equal since they also sit relative to the transverse line, OA)

Thus, AB || MN (QED)

8 0
2 years ago
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