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
dedylja [7]
3 years ago
10

I need help and I'm not sure if the answer is A so can you pls help!

Mathematics
1 answer:
adoni [48]3 years ago
6 0
The answer is not A because it less than or equal to and 1 is = 1. 
The answer is D because -5 is less than 1 and its not equal to 1.
You might be interested in
Is there an outliner? if so, what is it histogram
frutty [35]

Answer:

Usually, the presence of an outlier indicates some sort of problem. This can be a case which does not fit the model under study, or an error in measurement. Outliers are often easy to spot in histograms.

Step-by-step explanation:

It is similar to a Bar Chart, but a histogram groups numbers into ranges

3 0
3 years ago
Which expression is equivalent to 4 times the sum of 3 and x
vovikov84 [41]

Answer:

4(3+x)

Step-by-step explanation:

If any thing is in the brackets you need to do it first

so if you do 3 + x × 4 it will give yo 4x + 3 or 3 +4x

8 0
3 years ago
Solve for x use the completing the square method x^2+6x=5
Mrac [35]

Answer:

x_{1} =-3 +\sqrt{14} \\\\x_{2} =-3 -\sqrt{14}

Step-by-step explanation:

x^{2}+6x-5=0

we divide the coefficient of the X by half :

in this case: 6/2 = 3 , then we do the following

The result obtained is raised to square power:  3^2=9

we sum and subtract by 9 to maintain the balance of the equation:

x^{2}+6x+9-9-5=0

we have:

(x+3)^{2}-9-5=0

(x+3)^{2} = 14

lets apply square root on both sides of the equation:

\sqrt{(x+3)^{2}}=\sqrt{14}

we know:

\sqrt{a^{2}} = abs(a)

so we have:

abs(x+3)=\sqrt{14}

from where two solutions are obtained

x_{1} + 3 =\sqrt{14} \\\\x_{2} + 3 =-\sqrt{14}

finally we have:

x_{1} =-3 +\sqrt{14} \\\\x_{2} =-3 -\sqrt{14}

5 0
3 years ago
Read 2 more answers
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tr
frozen [14]

Answer:

The original height of the tree is 18 m.

Step-by-step explanation:

Please see attached photo for explanation.

From the diagram, we shall determine the value of 'x'. This can be obtained by using the pythagoras theory as follow:

x² = 5² + 12²

x² = 25 + 144

x² = 169

Take the square root of both side

x = √169

x = 13 m

Finally, we shall determine the original height of the tree. This can be obtained as follow.

From the question given above, the tree was broken from a height of 5 m from the ground which form a right angle triangle with x being the Hypothenus as illustrated in the diagram.

Thus, the original height of the will be the sum of 5 and x i.e

Height = 5 + x

x = 13 m

Height = 5 + 13

Height = 18 m

Therefore, the original height of the tree is 18 m.

3 0
3 years ago
Simplify (2+7) divided by (7-2 times 2)
Step2247 [10]
1.111.... so just right 1.2
3 0
3 years ago
Other questions:
  • Determine the missing side length:
    6·1 answer
  • How many different prisms can you make using 18 centimetre cubes?
    13·1 answer
  • Can someone answer this question please answer it correctly if it’s corect I will mark you brainliest
    7·2 answers
  • PLZZZZZZ NEED HELP!!!!!!!!
    9·1 answer
  • A rectangular rug has
    11·1 answer
  • BRAINLIEST ASAP! PLEASE HELP ME :)
    13·1 answer
  • Jaren estimated that she could run at a steady rate of 8km/h. She enters a marathon, a distance of 36 km. How long should it tak
    13·1 answer
  • I dropped my xbox conrtroller and one of the back buttons got locked up any help?
    10·1 answer
  • The length from the center of a wind turbine to the tip of one of its blades measures 32 meters. What is the approximate distanc
    15·1 answer
  • Which of the following are remote interior angles of <1??
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!