1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
gizmo_the_mogwai [7]
3 years ago
9

which of the following statements must be true about an equation before you can use the quadratic formula to find the solutionsl

Mathematics
1 answer:
Pepsi [2]3 years ago
8 0
I don't see anything? Perhaps you need a, b & c values to start
You might be interested in
What is the solution for 2t <5
Luba_88 [7]
Divide 2 from both sides
t < 5/2
Hope this helps!
6 0
3 years ago
Read 2 more answers
1. Yvonne closes her eyes and picks a candy from the bag, hoping for a blueberry. What are the chances she will pick a blueberry
elixir [45]

how many candy in the bag

5 0
3 years ago
A box with a square base and an open top is being constructed out of A cm2 of material. If the volume of the box is to be maximi
viktelen [127]

Answer:

Side length = \sqrt{\frac{A}{3} } cm ,   Height =  \frac{1}{2} \sqrt{\frac{A}{3} } cm  ,  Volume = \frac{A\sqrt{A}}{6\sqrt{3} }  cm³

Step-by-step explanation:

Assume

Side length of base = x

Height of box = y

total material required to construct box = A ( given in question)

So it can be written as

A = x² + 4xy

4xy = A - x²

  1. y = \frac{A - x^{2} }{4x}

Volume of box = Area x height

V = x² ₓ y

V = x² ₓ ( \frac{A - x^{2} }{4x} )

V =  \frac{Ax - x^{3} }{4}

To find max volume put V' = 0

So taking derivative equation becomes

\frac{A - 3 x^{2} }{4} = 0

A = 3 x^{2}

x^{2} = \frac{A}{3}

x = \sqrt{\frac{A}{3\\} }

put value of x in equation 1

y = \frac{A - \frac{A}{3} }{4\sqrt{\frac{A}{3} } }  

y = \frac{2 \sqrt{\frac{A}{3} } }{4 \sqrt{\frac{A}{3} } }

y = \frac{1}{2} \sqrt{\frac{A}{3} }

So the volume will be

V = x^{2} × y

Put values of x and y from equation 2 & 3

V = \frac{A}{3} (\frac{1}{2} \sqrt{\frac{A}{3} } )

V = \frac{A\sqrt{A}}{6\sqrt{3} }

8 0
4 years ago
Please help with the question,
castortr0y [4]
The correct answer is x^2-2x+2=0

 The explanation is shown below:

 1. To solve the problem shown in the figure above, you must apply the following proccedure:

 2. You have that the roots of the quadratic equation shown in the figure attached in the problem are:

  x=-1±i

 3.  Then, you know the quadratic formula to solve quadratic equations, which is:

 (-b±√(b^2-4ac))/2a

 4. You can see in the figure that a=1 and c=2; then, by analizing this, you can conclude that the coefficient b is:

 b=-2

 5. Therefore, you can conclude tha the quadratic equation is:

 x^2-2x+2=0

 6. If you want to verify it, apply the quadratic formula to the quadratic equation shown above and you will obtain the roots shown in the figure.
8 0
3 years ago
Read 2 more answers
HELP ASAP<br><br>what is the domain of the function shown on the graph?<br>​
fenix001 [56]
I think it’s A I’m not for sure tho but yeah the Domain is the x the first is -10 and I think the line is on 2 for the other x
4 0
3 years ago
Other questions:
  • every evening jenna empties her pockets and puts her change in a jar. At the end of the week she counts her money. One week she
    5·1 answer
  • The jogging track at Francine's school is 3/4 mile long. Yesterday Francine completed two laps on the track. If she ran 1/3 of t
    5·1 answer
  • 50 parents were asked which activities they enrolled their infant in.
    14·1 answer
  • Help please . Lost .
    9·1 answer
  • 6 + 3 = 9 and 5 · 5 = 20._____<br>True or False?
    5·2 answers
  • Find area of a square garden having a length 45m ​
    5·2 answers
  • Can someone please help?
    14·1 answer
  • Arti bought 12 kilogram of sweets. How many<br> grams did she buy?
    5·1 answer
  • PLZ HELP ME ILL MARK BRAINLIETS
    10·1 answer
  • The sides of a square are 12 feet long. Find the new side<br> length after it is increased by 40%.
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!