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
34kurt
3 years ago
7

state whether the following statement is true or false: y= -x2 + 6x - 10 does not intersect the x-axis.​

Mathematics
1 answer:
oee [108]3 years ago
4 0

Answer:

True

Step-by-step explanation:

Here, we want to check if the given equation does not intersect the x-axis

if it does not, then it has no real root

If the discriminant is less than zero, then it has no real roots and does not intersect the x-axis

The formula for the discriminant is;

D = b^2-4ac

In the question;

a = -1 , b = 6 , c = 10

so;

D = 6^2-4(-1)(10)

D = 36-40

D = -4

Since discriminant is less than zero, then there is no real root and the graph does not cross the x-axis

You might be interested in
Math equasion that equals 8
ICE Princess25 [194]

If there aren't any specific requirements for the equation, here are a few choices:

2³ = 8

64 / 8 = 8

2x + 4 = y (where x = 2)

x² - 1 = y (where x = 3)

6 0
3 years ago
21% of the students at a High are in Maths group, 15 % are in science group, and 6 % are in both. If a student is selected rando
mihalych1998 [28]

Answer:

6%

Step-by-step explanation:

The probability of picking a maths student = 21% = 21/100

The probability of picking a science student = 15% = 15/100 = 3/20

The probability of picking in both = 6/100 = 3/50

The probability that the student is in maths or science = 21/100 - 15/100 = 6/100 = 6%.

4 0
2 years ago
Simplify the equation <br><br> 4(-8x + 5) - (-33x - 26)= ???
kkurt [141]

- 16 x+ 20 + 33x  + 26 \\ 17x + 46
4 0
3 years ago
Help me please really help
Phantasy [73]

Answer:

60

Step-by-step explanation:

10 x sqrt64 - 5 x 4

10 x 8 - 5 x 4

80 - 20

60

5 0
2 years ago
M men and w women seat themselves at random in m+w seats arranged in a row. find the probability that all the women will be adja
WINSTONCH [101]

The <em>correct answer</em> is:


\frac{(m+1)!w!}{(m+w)!}


Explanation:


We want all <em>w</em> women to be seated together. There are <em>w</em>! ways to do this.


Since all women are seated together, we consider the as 1 block to be seated with the men.


There are <em>m</em>! ways of arranging the men. However, we also have the 1 block of women to seat; this makes (<em>m</em>+1)! ways to seat the men and block of women.


There are (<em>m</em>+<em>w</em>)! ways to arrange all of the men and women.


This makes our probability

\frac{(m+1)!w!}{(m+w)!}.


For example, if there are 4 men and 3 women:

There are 3! = 6 ways to seat the women together. This makes 1 block of women.


There are 4! = 24 ways to seat the men together. Taking this with the block of women, we have (4+1)! = 5! = 120 ways to seat the men and block of women.


There are (4+3)! = 7! = 5040 ways to arrange 7 people.


This makes our probability 120(6)/5040 = 720/5040.

8 0
3 years ago
Other questions:
  • Payton plays 15 games of fortnite in 15 minutes. If he maintains this rate, how many games will he play in 3 hours
    12·2 answers
  • HOW TO SOLVE 2/5x=-25
    15·2 answers
  • Help needed!!! Serious answers only
    14·1 answer
  • a bicycle store costs $2400 per month to operate. the store pays and average of $60 per bike. the average selling price of each
    7·1 answer
  • 2. What is the slope passing through the points (2, -1) and (-4, 11)
    10·1 answer
  • Urgent, It is a Calculus question and I’ll appreciate your help. Thanks
    11·1 answer
  • You and your friends are having a race, but since you are faster than he is, you decide to give him a head start. He starts 10 m
    15·1 answer
  • Solve <br>60000000+63636733-7373 ​
    8·1 answer
  • Please help asap asap
    11·2 answers
  • This probability distribution shows the
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!