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Klio2033 [76]
4 years ago
13

Which of the following is not equivalent to the others?

Mathematics
1 answer:
maks197457 [2]4 years ago
7 0
C. 10/80 = 1/8 = 0.125
You might be interested in
Determine the equation of the circle graphed below. <br><br> ( please help me )
Brums [2.3K]

Answer:

Equation : \\(x+3)^2 + ( y -2)^2 = 36

Step-by-step explanation:

For standard form the circle's equation we need the centre of the circle and the radius.

Step 1: <em><u>Find the centre</u></em>

If the centre is not given find the end points of the diameter

and then find the mid point.

Let the end points of the diameter be : ( - 3 , 8 ) and ( -3 , -4 )

The mid-point of the diameter is :

                        Mid-point = (\frac{-3 + - 3}{2}, \frac{-4+8}{2}) =  (-3, 2)

Therefore, centre of the circle = ( -3 , 2 )

Step 2 : <u>Find radius</u>

<u></u>Radius = \frac{Diameter }{2}<u></u>

Diameter is the distance between the end points ( -3 , 8) and ( -3 , -4 )

That is ,

 Diameter = \sqrt{(-3-(-3))^2 + ( -4 -8)^2}\\

                 = \sqrt{(-3 + 3)^2 + (-12)^2}\\\\=\sqrt{0 + 144}\\\\=12

Therefore ,

       Radius = \frac{12}{2} = 6

Step 3 : <u>Equation of the circle</u>

Standard equation of the circle with centre ( h ,k )

and radius ,r is :

       (x - h)^2+(y -k)^2 = r^2

Therefore, the equation of the circle with centre ( -3, 2)

and radius = 6 is :

    (x - (-3))^2 +  (y - 2)^2 = 6^2\\\\(x + 3)^2 + (y - 2)^2 = 36

5 0
3 years ago
How do you complete the other two?
Gwar [14]

For now, I'll focus on the figure in the bottom left.

Mark the point E at the base of the dashed line. So point E is on segment AB.

If you apply the pythagorean theorem for triangle ABC, you'll find that the hypotenuse is

a^2+b^2 = c^2

c = sqrt(a^2+b^2)

c = sqrt((8.4)^2+(8.4)^2)

c = 11.879393923934

which is approximate. Squaring both sides gets us to

c^2 = 141.12

So we know that AB = 11.879393923934 approximately which leads to (AB)^2 = 141.12

------------------------------------

Now focus on triangle CEB. This is a right triangle with legs CE and EB, and hypotenuse CB.

EB is half that of AB, so EB is roughly AB/2 = (11.879393923934)/2 = 5.939696961967 units long. This squares to 35.28

In short, (EB)^2 = 35.28 exactly. Also, (CB)^2 = (8.4)^2 = 70.56

Applying another round of pythagorean theorem gets us

a^2+b^2 = c^2

b = sqrt(c^2 - a^2)

CE = sqrt( (CB)^2 - (EB)^2 )

CE = sqrt( 70.56 - 35.28 )

CE = 5.939696961967

It turns out that CE and EB are the same length, ie triangle CEB is isosceles. This is because triangle ABC isosceles as well.

Notice how CB = CE*sqrt(2) and how CB = EB*sqrt(2)

------------------------------------

Now let's focus on triangle CED

We just found that CE = 5.939696961967 is one of the legs. We know that CD = 8.4 based on what the diagram says.

We'll use the pythagorean theorem once more

c = sqrt(a^2 + b^2)

ED = sqrt( (CE)^2 + (CD)^2 )

ED = sqrt( 35.28 + 70.56 )

ED = 10.2878569196893

This rounds to 10.3 when rounding to one decimal place (aka nearest tenth).

<h3>Answer: 10.3</h3>

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

Now I'm moving onto the figure in the bottom right corner.

Draw a segment connecting B to D. Focus on triangle BCD.

We have the two legs BC = 3.7 and CD = 3.7, and we need to find the length of the hypotenuse BD.

Like before, we'll turn to the pythagorean theorem.

a^2 + b^2 = c^2

c = sqrt( a^2 + b^2 )

BD = sqrt( (BC)^2 + (CD)^2 )

BD = sqrt( (3.7)^2 + (3.7)^2 )

BD = 5.23259018078046

Which is approximate. If we squared both sides, then we would get (BD)^2 = 27.38 which will be useful in the next round of pythagorean theorem as discussed below. This time however, we'll focus on triangle BDE

a^2 + b^2 = c^2

b = sqrt( c^2 - a^2 )

ED = sqrt( (EB)^2 - (BD)^2 )

x = sqrt( (5.9)^2 - (5.23259018078046)^2 )

x = sqrt( 34.81 - 27.38 )

x = sqrt( 7.43 )

x = 2.7258026340878

x = 2.7

--------------------------

As an alternative, we could use the 3D version of the pythagorean theorem (similar to what you did in the first problem in the upper left corner)

The 3D version of the pythagorean theorem is

a^2 + b^2 + c^2 = d^2

where a,b,c are the sides of the 3D block and d is the length of the diagonal. In this case, a = 3.7, b = 3.7, c = x, d = 5.9

So we get the following

a^2 + b^2 + c^2 = d^2

c = sqrt( d^2 - a^2 - b^2 )

x = sqrt( (5.9)^2 - (3.7)^2 - (3.7)^2 )

x = 2.7258026340878

x = 2.7

Either way, we get the same result as before. While this method is shorter, I think it's not as convincing to see why it works compared to breaking it down as done in the previous section.

<h3>Answer:  2.7</h3>
8 0
3 years ago
Without solving the equation, determine the nature of the roots of x2 - 2 x - 8 = 0.
ad-work [718]
Evaluate using the discriminant (the part of the quadratic equation under the square root sign) b²-4ac
In your equation, a=1, b=(-2), and c=(-8)...substitute their values into the discriminant:
(-2)²-4(1)(-8)
4+32 = 36

The value is positive, therefore the equation has two real roots.
It's a perfect square (√36=6) so the roots are rational
Answer: two real, rational roots

(If the result was a negative number, the system would have two irrational roots. If the result was zero, the system would have a <em>double root </em>(one value repeated twice - when graphed, the vertex of the parabola falls on the x-axis)∨<em />
3 0
4 years ago
Leo is drinking tea . he drinks 3 1/2 liter of tea every 2/3 of an hour . how many liters of tea he drinks in one hour?
My name is Ann [436]
Divide 3 1/2 by 2/3.

3 1/2 = 7/2 as an improper fraction, so we have

7/2 ÷ 2/3

= 7/2 * 3/2

= 21/4 = 5 1/4 L
5 0
3 years ago
Help please. Middle school math
Zanzabum

Answer:

  • 8 - H
  • 9 - T
  • 10 - D
  • GARFIELD THE CAT

Step-by-step explanation:

Correct so far.

  H -- 8

  T -- 9

  D -- 10

The balloon depicted ...

  GARFIELD THE CAT

_____

The applicable rules of exponents are ...

  (a^b)(a^c) = a^(b+c)

  (a^b)^c = a^(b·c)

Of course, an odd number of minus signs in a product means the product is negative.

__

Most of these can be figured just by looking at the coefficient, or the exponent of one variable, such as x.

8 0
3 years ago
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