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
Aliun [14]
3 years ago
14

Replace ... with ≥ or ≤ so that the derived inequality will be true for any value for x: x^2-16x+64 ... 0

Mathematics
1 answer:
ankoles [38]3 years ago
8 0

Answer:

≥

..................................................

You might be interested in
Help help help please!!!!!!!!!!
Paul [167]

Answer:

{x}^{2}  + 2x + 1

Step-by-step explanation:

They ask: f(x) - g(x) =

( {x}^{2}  + 3x + 2) - (x + 1)

{x}^{2}  + 2x + 1

5 0
3 years ago
Help me please thank you
TEA [102]

Answer:

about 41.14 pages per day

Step-by-step explanation:

She wants to read 576 pages in 14 days

Take the number of pages and divide by the number of days

576/14=41.14285714 pages per day


6 0
3 years ago
Read 2 more answers
If tanA+sinA=m and tanA-sinA=n.prove that m^2-n^2=4√mn
mash [69]

Let a=\tan A and b=\sin A. Then

m^2-n^2=(a+b)^2-(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)=4ab

\implies m^2-n^2=4\tan A\sin A

and

mn=(a+b)(a-b)=a^2-b^2

\implies4\sqrt{mn}=4\sqrt{\tan^2A-\sin^2A}

The expression under the square root can be rewritten as

\tan^2A-\sin^2A=\dfrac{\sin^2A}{\cos^2A}-\sin^2A=\sin^2A\left(\dfrac1{\cos^2A}-1\right)=\sin^2A(\sec^2A-1)

Recall that

\sin^2A+\cos^2A=1\implies\tan^2A+1=\sec^2A

so that

\tan^2A-\sin^2A=\sin^2A\tan^2A

and assuming \sin A>0 and \tan A>0, we end up with

4\sqrt{\tan^2A-\sin^2A}=4\tan A\sin A

so that

m^2-n^2=4\sqrt{mn}

as required.

5 0
3 years ago
Use a unit​ circle, a 30 degrees - 60 degrees - 90 degrees ​triangle, and an inverse function to find the degree measure of the
PolarNik [594]

150 is the answer I don’t think it is right

8 0
3 years ago
2k^2-5k-18=0 what is both of the values of k? plz help!!! 15 pts!!!
ddd [48]
To factor quadratic equations of the form ax^2+bx+c=y, you must find two values, j and k, which satisfy two conditions.

jk=ac and j+k=b

The you replace the single linear term bx with jx and kx. Finally then you factor the first pair of terms and the second pair of terms. In this problem...

2k^2-5k-18=0

2k^2+4k-9k-18=0

2k(k+2)-9(k+2)=0

(2k-9)(k+2)=0

so k=-2 and 9/2

k=(-2, 4.5)
5 0
3 years ago
Other questions:
  • HELP!!! 10 points and marked brainliest
    13·1 answer
  • 3.
    15·2 answers
  • What is the slope of the line? 8x-6y=1<br><br> also, thanks
    13·1 answer
  • What inequality represents the verbal expression? all real numbers less than 69
    14·2 answers
  • What's the answer and how do you do it?
    13·2 answers
  • How much material is needed to make a triangular flag with base 2 1/4 and height 8 1/2
    13·1 answer
  • Round 462 to the nearest 100.
    5·2 answers
  • Transformation is given as (x, y) -&gt;x + 3, y + 6), Find the
    8·1 answer
  • Subtract (3x - 7x^2 + 2) - (4x^2 - 5 + 6x)
    12·1 answer
  • Have you ever seen a photo online that did not seem to have the colors right? Perhaps the colors looked unnatural or not appropr
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!