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ICE Princess25 [194]
2 years ago
7

[tex]\sqrt{49} please answer correctly

Mathematics
1 answer:
Nookie1986 [14]2 years ago
7 0

Answer:

Um.. well I don't know what is your question, but I think you meant square root and the answer is 7 because it is a perfect square.

Step-by-step explanation:

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A consumer report counted 72 low-quality, 112 medium-quality, and 711 high-quality computers based on a random sample. The repor
blsea [12.9K]

Answer:

high quality = 711/895

medium quality = 112/895

low quality = 72/895

Step-by-step explanation:

72 + 112 + 711 = 895

high quality = 711/895

medium quality = 112/895

low quality = 72/895

8 0
2 years ago
I need help with 3-4 quick
Agata [3.3K]
3-4 = -1

Hope this helps!

:)


6 0
3 years ago
What is a necessary step for constructing perpendicular lines through a point off the line?
Nostrana [21]

Answer:

Find another point on the perpendicular line.

Step-by-step explanation:

Given an original line "m", and a point off the line "Q", in order to construct a second line "p", meant to be perpendicular to "m" through the point "Q", fundamentally, the only truly necessary step to construct a perpendicular line through is to find another point on the yet-to-be-found perpendicular line.

Most often, this is accomplished by exploiting the fact that "p" is the set of all points that are equidistant from any pair of points that are symmetric about "p".

Since the symmetry must be about "p", and we don't even know where "p" is, one often finds two points on "m" that are equidistant from "Q".

This can be accomplished by adjusting a compass to a fixed radius (larger than the distance from "Q" to "m"), and making an arc that intersects "m" in two places.  Those two places will be equidistant from "Q", and are simultaneously on line "m".  Thus, these two points, "A" & "B" are symmetric about "p".

Since "A" & "B" are symmetric about "p", they are equidistant from "p", and are on "m".  One could try to find the point of intersection between "p" and "m" through construction, but this is unnecessary.  We need only find a second point (besides "Q") that is equidistant from "A" & "B", which will necessarily be a point on "p", to form the line perpendicular to "m".

To do this, fix the compass with any radius, and from "A" make a large arc generally in the direction of "B", and make the same radius arc from "B" in the direction of "A" such that the two arcs intersect at some point that isn't "Q".  This point of intersection we can call point "T", and the line QT is line "p", the line perpendicular to the original line, necessarily containing "Q".

8 0
2 years ago
Make a the subject of a2+b2=c2
wel

Answer:

a=\sqrt{c^2-b^2

⇒ a=\sqrt{(c+b)(c-b)}        [Factorized form]

Step-by-step explanation:

Given expression:

a^2+b^2=c^2

We need to make a the subject.

In order to do that we need to solve the expression for a in terms of b and c

We have,

a^2+b^2=c^2

Subtracting both sides by b^2

a^2+b^2-b^2=c^2-b^2

a^2=c^2-b^2

Taking square root both sides in order to remove square of a

\sqrt{a^2}=\sqrt{c^2-b^2}

a=\sqrt{c^2-b^2

The difference of squares can be factorized and written  as:

a=\sqrt{(c+b)(c-b)}    [difference of squares x^2-y^2=(x+y)(x-y) ]

4 0
3 years ago
Rectangles Are a subset of which set
Slav-nsk [51]
They are a subset to squares
4 0
3 years ago
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