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2 years ago

Is it possible to square root a negative number?

2 answers:

3 0

<span>No because a negative times a negative will be positive and it will not equal a negative number. the square root cannot be one negative and one positive because that is against the definition of a square root.
</span>

Taya2010 [7]2 years ago

3 0

No because you had no number which were negative after you squared. if you do every number goes back positive after you squared it i think.

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A sector with an area of \goldE{48\pi,\text{cm}^2}48πcm 2 start color #a75a05, 48, pi, start text, c, m, end text, squared, end

erik [133]

Answer:

3/8 π radians

Step-by-step explanation:

The Area of a sector when then central angle is in radians = 1/2r² θ

Where

θ = central angle = ?

r = 16 cm

Area of the sector = 48πcm²

Hence

Central angle = Area of a sector ÷ (1/2r²)

= 48πcm² ÷ (1/2 × 16²)

= 48πcm² ÷ 128

Central angle = 3/8π radians

Therefore, Central angle = 3/8π radians

7 0

2 years ago

Can someone help me

Mashcka [7]

Answer:

whois someone? OH ! Right, That dude named Some Won . Sorry.. i'll let him answer.

Step-by-step explanation:

8 0

2 years ago

Plz help it's due tomorrow

posledela

Most of the answers are in the book. 1&2 is based on vocabulary and 8&9 are on the chart

7 0

2 years ago

What is the gradient of the graph shown?<br> Give your answer in its simplest form.

Amiraneli [1.4K]

Answer:

-2

Step-by-step explanation:

I think this was the question you were talking about.

4 0

2 years ago

Triangle MNO is an equilateral triangle with sides measuring 16 StartRoot 3 EndRoot units. Triangle M N O is an equilateral tria

Lesechka [4]

Answer:

The correct option is second one i.e 24 units.

Therefore the height of the triangle is

$NR=24\ units$

Step-by-step explanation:

Given:

An equilateral triangle has all sides equal.

ΔMNO is an Equilateral Triangle with sides measuring,

$NM = MO = ON =16\sqrt{3}$

NR is perpendicular bisector to MO such that

$MR=RO=\dfrac{MO}{2}=\dfrac{16\sqrt{3}}{2}=8\sqrt{3}$ .NR ⊥ Bisector.

To Find:

Height of the triangle = NR = ?

Solution :

Now we have a right angled triangle NRM at ∠R =90°,

So by applying Pythagoras theorem we get

$(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}$

Substituting the values we get

$(MN)^{2} = (MR)^{2}+(NR)^{2}\\\\(16\sqrt{3})^{2}=(8\sqrt{3})^{2}+(NR)^{2}\\\\(NR)^{2}=768-192=576\\Square\ rooting\ we\ get\\NR=\sqrt{576}=24\ units$

Therefore the height of the triangle is

$NR=24\ units$

6 0

2 years ago

Read 2 more answers