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docker41 [41]
3 years ago
10

The area of a square is 4 square meters. How long is each side? Please show every step.

Mathematics
2 answers:
inessss [21]3 years ago
3 0
Since you're trying to find the side lenghts of a square, all we need to do is square root the area. So, the square root of 4 (the area) is 2.

So, all of the sides of the square are 2 meters long.
Rom4ik [11]3 years ago
3 0
Since the area of the square is given by side length squared....

let s=side length
s²=4
s=±√4 or±2

but since length can never be negative, we ignore there extraneous answer-2 and conclude the side length s=2
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HELP ASAP!! LOTS OF POINTS!! WILL MARK BRIAN-LIST
KiRa [710]

Answer:

LSA = 532 yds ^2

Step-by-step explanation:

We do not add the triangles in because they are the bases and the bases do not get added in the lateral surface areas.

From left to right

Rectangle 1

A = lw = 9.9 *20 =198

Rectangle 2

A = lw = 6.8 *20 =136

Rectangle 3

A = lw = 9.9 *20 =198

Add them together

198+136+198

532

LSA = 532 yds ^2

7 0
3 years ago
Find the greatest common factor. Show work!
zvonat [6]

Answer:

4y

Step-by-step explanation:

Because it is the GCF of the numbers and it can't have an exponent since 16y has none.

3 0
3 years ago
In my town, gas prices are always listed to the thousandths place. Since the smallest coin we have is
Virty [35]

Answer:

It would be rounded to $5.90 per gallon. As if something is 5 or above you round up and if it is 4 or less you round down.

8 0
3 years ago
I'm being timed pls help will pick brainliest
stiv31 [10]

Answer:

Cos 0° = 6.42 = V1

6th quadrant.

So, cos x = 0 implies x = (2n + 1)π/2 , where n takes the value of any integer. For a triangle, ABC having the sides a, b, and c opposite the angles A, B, and C, the cosine law is defined. In the same way, we can derive other values of cos degrees like 30°, 45°, 60°, 90°, 180°, 270°and 360

Step-by-step explanation:

if sin a= 3/4 then a = 50

if sin = 4/5 then a = 60

if sin = 2/3 then a = 40

But we can perfect this

if sin = 4/5 then a = 57   as 4/5 = 0.83

We want 0.8 = 4/5

if sin = 4/5 then a = 54 as 4/5 = 0.809

if sin = 4/5 then a = 53.5 as 4/5 = 0.80385

Now for cos

It is much easier than it initially appears. Remember the definition of SINE:

SINΘ = opp             In your case, that means the opposite is 4/5 = 0.80385 (yes, ignore the sign for now) and the hypotenuse is 11.48910018

           hyp             Please draw that triangle right now, because it will help you a lot at the end.

 53.50 degree           Remember to place the angle in the appropriate spot.

Now, use the Pythagorean Theorem to find the missing side (easy, right? It's 9.534) and place it in the adjacent position.

You can easily find all of the trig functions now!

Simply remember that:

COSΘ = Adj                                  with SEC the reciprocal of this one

            Hyp

TANΘ = Opp                                  COT the reciprocal of TAN and, if anyone asks, CSC coming from SIN.

            Adj

I told you to ignore the signs, but now we can't anymore. Remember the four quadrants and the memory trick:

A                        -- ALL are positive

Smart                 -- SIN and CSC are positive

Trig                    -- TAN & COT are positive

Class                  -- COS and SEC are positive.

Since your SIN was negative, it must be in III or 6, and COS is positive in I and 6 So we're in quadrant 6 then!

Only your COS and SEC will be positive, the rest negative

7 0
3 years ago
Suppose you choose a team of two people from a group of n > 1 people, and your opponent does the same (your choices are allow
jonny [76]

Answer:

The number of possible choices of my team and the opponents team is

 \left\begin{array}{ccc}n-1\\E\\n=1\end{array}\right     i^{3}

Step-by-step explanation:

selecting the first team from n people we have \left(\begin{array}{ccc}n\\1\\\end{array}\right)  = n possibility and choosing second team from the rest of n-1 people we have \left(\begin{array}{ccc}n-1\\1\\\end{array}\right) = n-1

As { A, B} = {B , A}

Therefore, the total possibility is \frac{n(n-1)}{2}

Since our choices are allowed to overlap, the second team is \frac{n(n-1)}{2}

Possibility of choosing both teams will be

\frac{n(n-1)}{2}  *  \frac{n(n-1)}{2}  \\\\= [\frac{n(n-1)}{2}] ^{2}

We now have the formula

1³ + 2³ + ........... + n³ =[\frac{n(n+1)}{2}] ^{2}

1³ + 2³ + ............ + (n-1)³ = [x^{2} \frac{n(n-1)}{2}] ^{2}

=\left[\begin{array}{ccc}n-1\\E\\i=1\end{array}\right] =   [\frac{n(n-1)}{2}]^{3}

4 0
3 years ago
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