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Serhud [2]
3 years ago
15

A student performed the following steps to find the solution to the equation x^2+14x+40=0 where did the student go wrong

Mathematics
2 answers:
EastWind [94]3 years ago
7 0
I don't fully know exactly what your asking and I'm new at this, but I think you're asking about what is the center and radius of a circle right? In that case, this is what I came up with (Don't know if this is what you want, sorry) :(
First step: You do -40 to the current +40, and transfer it to the other side of the equal symbol, making the 0 turn into a -40.
Second step: It should now look like this - (x^2+14x) = -40
Third step: Divide the 14 by 2, which gives you a positive 7, then square that 7, which gives you 49. It shall look like this now - (x^2+14x+49) = 9. The nine is now there at the end because you add the 49 to the negative 40.
Fourth step: Remember the 7? Well, we use it now. It should currently look like this - (x+7) = 9. 
Fifth step: The Center can now be found, and remember for the center you do the opposite of what the number is, in this case, the positive 7, will be a negative 7 in the center. Since there is no y found, it will remain as 0. (I think) This is how It looks - Center: (-7, 0)
Sixth step (Final step): Now the radius must be found, we will now look at the nine at the other side of the equal symbol. We square root that now! So the radius = 3. 
Answers: Center: (-7, 0) Radius: 3. (Or if asked to find diameter, then it would be double the 3. Meaning Diameter=6.
Comment: I'm only a 9th grader, so sorry If I got this wrong. Trying to help :p lol. 
Yanka [14]3 years ago
7 0

Answer:

It was in step 3

Step-by-step explanation: APEX

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The sum of two numbers is 90. Their difference is 18. Find<br> the numbers.
Naddik [55]

Answer:

36 & 54

Step-by-step explanation:

<em><u>By </u></em><em><u>Question</u></em><em><u> </u></em><em><u>:</u></em><em><u>-</u></em><em><u> </u></em>

  • Sum of numbers = 90 .
  • Difference of numbers = 18 .

<em><u>Let </u></em><em><u>:</u></em><em><u>-</u></em><em><u> </u></em>

  • First Number be x .
  • Second number be y .

<em><u>According</u></em><em><u> to</u></em><em><u> </u></em><em><u>1</u></em><em><u>s</u></em><em><u>t</u></em><em><u> </u></em><em><u>condition</u></em><em><u> </u></em><em><u>:</u></em><em><u>-</u></em><em><u> </u></em>

: \implies x + y = 90

<u>According</u><u> to</u><u> </u><u>2</u><u>n</u><u>d</u><u> </u><u>condition</u><u> </u><u>:</u><u>-</u><u> </u>

: \implies x - y = 18

<em><u>Adding</u></em><em><u> the</u></em><em><u> </u></em><em><u>equations</u></em><em><u> </u></em><em><u>:</u></em><em><u>-</u></em><em><u> </u></em>

: \implies 2x = 18 + 90

: \implies 2x = 108

: \implies x = 54

<em><u>Therefore</u></em><em><u> </u></em><em><u>:</u></em><em><u>-</u></em><em><u> </u></em>

: \implies First number = 54

: \implies Second number = 36

5 0
3 years ago
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Please help me answer this question
stealth61 [152]

The value of dy/dx for the functions are

(i) \frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x

(ii) \frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }

<h3>Differentiation</h3>

From the question, we are to determine dy/dx for the given functions

(i) x^{2} sin^{2}2x

Let u = x^{2}

and

v = sin^{2} 2x

Also,

Let w=  sin2x

∴ v = w^{2}

First, we will determine \frac{dv}{dx}

Using the Chain rule
\frac{dv}{dx} = \frac{dv}{dw}.\frac{dw}{dx}

v = w^{2}

∴ \frac{dv}{dw} =2w

Also,

w=  sin2x

∴ \frac{dw}{dx} =2cos2x

Thus,

\frac{dv}{dx} = 2w \times 2cos2x

\frac{dv}{dx} = 2sin2x \times 2cos2x

\frac{dv}{dx} = 4sin2x . cos2x

Now, using the product rule

\frac{dy}{dx} = u\frac{dv}{dx} +  v\frac{du}{dx}

From above

u = x^{2}

∴ \frac{du}{dx}=2x

Now,

\frac{dy}{dx} = x^{2} (4sin2x.cos2x)+  sin^{2}2x (2x)

\frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x

(ii) xy^{2}+y^{2}x^{3} +2=0

Then,

x.2y\frac{dy}{dx}+ y^{2}(1)+y^{2}.3x^{2} + x^{3}.2y\frac{dy}{dx} +0=0

2xy\frac{dy}{dx}+ y^{2}+3x^{2}y^{2} + 2x^{3}y\frac{dy}{dx} =0

2xy\frac{dy}{dx}+2x^{3}y\frac{dy}{dx} =-  y^{2}-3x^{2}y^{2}

\frac{dy}{dx} (2xy+2x^{3}y)=-  y^{2}(1+3x^{2})

\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy+2x^{3}y}

\frac{dy}{dx} =\frac{- y^{2}(1+3x^{2})}{2xy(1+x^{2}) }

\frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }

Hence, the value of dy/dx for the functions are

(i) \frac{dy}{dx} = 4x^{2}sin2x.cos2x+ 2x. sin^{2}2x

(ii) \frac{dy}{dx} =\frac{- y(1+3x^{2})}{2x(1+x^{2}) }

Learn more on Differentiation here: brainly.com/question/24024883

#SPJ1

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Ms. Anastasopoulos has 30 students. 4/5 of the students have pets. OF the students who have pets, 2/3 of them have a dog as a pe
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Answer:

16

Step-by-step explanation:

since 4/5 have a pet, you would divide by 5.

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but since only 4 have a pet, you would multiply that by 4 to get 4/5

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now you would divide by 3, as 2/3 have a dog

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but only 2/3 have dog, so you would multiply by 2

8×2=16

I'm sure there is a simpler way but this is how I would do it.

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