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eduard
3 years ago
10

What is the area of this square?

Mathematics
2 answers:
lbvjy [14]3 years ago
7 0

The formula for finding area is l * w

Since a square has all sides the same length, we need to multiply 40 by 40.


40 * 40 = 1,600 ←

Check work by dividing 1,600 and 40:

1,600 ÷ 40 = 40

<u>Your answer is Option D. 1,600 km². (The answer wouldn't be option C as it is referring to perimeter, not area)</u>


<u>The area of the square is 1,600 km².</u>


I hope this helps! :)

shepuryov [24]3 years ago
5 0
Area = 40x40 = 1600<span>km²

answer 

</span><span>D.
1,600 km²</span>
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The cell phone company will add −$2.74 to your next bill foreach of the 4 months you were overcharged.Will your next bill increa
velikii [3]

Answer:

It will be

Step-by-step explanation:

5 0
3 years ago
Express square root -144 in its simplest terms.​
Montano1993 [528]

Answer:

\sqrt{144} = 12 or  \sqrt{-144} = 12i

Step-by-step explanation:

You can't take squareroot of a negative number. It's not allowed!

But for positive 144 if that's what you're asking, it will be 12.

Hope this helps!

3 0
3 years ago
What is the equation of the following graph in vertex form?
Zepler [3.9K]
<h3>Answer: Choice B.   y = (x+3)^2 - 1</h3>

============================================

How I got that answer:

The vertex is the lowest point of parabolas that open upward.

(h,k) = vertex

(h,k) = (-3,-1)

h = -3

k = -1

For each of the answer choices, a = 1.

The general template of a quadratic in vertex form is

y = a(x-h)^2 + k

Plug a = 1, h = -3, k = -1 into that equation. Simplify.

y = a(x-h)^2 + k

y = 1(x-(-3))^2 + (-1)

y = (x+3)^2 - 1

8 0
3 years ago
Consider the parabola r​(t)equalsleft angle at squared plus 1 comma t right angle​, for minusinfinityless thantless thaninfinity
kodGreya [7K]

Given:-   r(t)=< at^2+1,t>  ; -\infty < t< \infty , where a is any positive real number.

Consider the helix parabolic equation :  

                                              r(t)=< at^2+1,t>

now, take the derivatives we get;

                                            r{}'(t)=

As, we know that two vectors are orthogonal if their dot product is zero.

Here,  r(t) and r{}'(t)  are orthogonal i.e,   r\cdot r{}'=0

Therefore, we have ,

                                  < at^2+1,t>\cdot < 2at,1>=0

< at^2+1,t>\cdot < 2at,1>=

                                              =2a^2t^3+2at+t

2a^2t^3+2at+t=0

take t common in above equation we get,

t\cdot \left (2a^2t^2+2a+1\right )=0

⇒t=0 or 2a^2t^2+2a+1=0

To find the solution for t;

take 2a^2t^2+2a+1=0

The numberD = b^2 -4ac determined from the coefficients of the equation ax^2 + bx + c = 0.

The determinant D=0-4(2a^2)(2a+1)=-8a^2\cdot(2a+1)

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola  r(t)=< at^2+1,t> i.e <1,0>




                                               




6 0
3 years ago
F(x) = 5x + 2x^2-7x+4<br> g(x) =3x-7 <br> find (f+g)(x)
Angelina_Jolie [31]

Answer:

(f+g)(x)=5x+2x²-7x+4+3x-7 = 2x²+x-3

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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