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
Kryger [21]
4 years ago
14

Which number line shows the solution to y <6

Mathematics
1 answer:
Georgia [21]4 years ago
7 0

Answer:

On the number line, the y would have to be 5 or bellow.

Step-by-step explanation:

You might be interested in
9.25 multiplied by 8.3
Montano1993 [528]

Answer:

76.775

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
How do i find the standard deviation of test scores
user100 [1]

Answer:

Step-by-step explanation:

Solve for the mean (average) of the five test scores.

Subtract that mean from each of the five original test scores. Square each of the differences.

Find the mean (average) of each of these differences you found in Step 2.

Take the square root of this final mean from #3. This is the standard deviation.

4 0
3 years ago
Read 2 more answers
if $20.90 in interest is earned on a 2 year investment at 5.5% a year, what amount was originally deposited?
andrezito [222]
You shall check the type of interest

7 0
4 years ago
A farmer has two pieces of land in a parallelogram shape on the same base. He grows vegetables and flowers in two triangular pie
Hitman42 [59]

Step-by-step explanation:

In trapezoid ABCF, AB || CF.

Therefore, parallelograms ABEF & ABCD have same height and lie on the same base AB.

\therefore  Ar(\parallel^{gm} ABEF) = Ar(\parallel^{gm}ABCD)\\... (1)\\\\   \because\: Ar(\parallel^{gm} ABEF) \\=Ar(\triangle AFD) + Ar(\square \: ABED)....(2)  \\\\\&\: Ar(\parallel^{gm}ABCD)\\= Ar(\triangle BCE) +Ar(\square ABED)....(3)

From equations (1), (2) & (3), we find:

Ar(\triangle AFD) + Ar(\square \: ABED)  \\ = Ar(\triangle BCE) +Ar(\square ABED)  \\  \\ \purple {\boxed {\therefore  Ar(\triangle AFD)    = Ar(\triangle BCE)}}  \\

Thus, the two triangular pieces of land are equal in area.

Hence proved.

8 0
3 years ago
Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. She bikes at a uniform rate of 12 miles per hour. Brian lea
exis [7]

Answer:

(D) 11: 00

Step-by-step explanation:

We have been given that Cassie leaves Escanaba at 8:30 AM heading for Marquette on her bike. Brian leaves Marquette at 9:00 AM heading for Escanaba on his bike.

This means that Cassie leaves 1/2 hour before Brian.

Since Cassie bikes at a uniform rate of 12 miles per hour, so distance covered in half hour would be half of 12 miles that is 6 miles.

Since total distance is 62 miles, so distance left between both at 9:00 Am would be 62-6=56.

We are told that Brian bikes at a uniform rate of 16 miles per hour and Cassie bikes at a uniform rate of 12 miles per hour, so distance traveled by both in one hour would be 16+12=28 miles.

Let us divide 56 by 28 to find time taken to complete 56 miles.

\frac{56}{28}=2.

Since it will take them 2 hours to cover 56 miles, therefore, they will meet 2 hours after 9:00 AM that is 11:00 AM.

3 0
4 years ago
Other questions:
  • Solve for xxx.
    13·1 answer
  • In equilateral triangle RST, R has coordinates (0, 0) and T has coordinates of (2a, 0). Find the coordinates of S in terms of a.
    5·2 answers
  • Quick help.<br> Will Mark as Brainliest if my laptop stops acting up.
    6·1 answer
  • -8
    8·1 answer
  • I need to find the greatest common factor for these numbers and what GCF do they have in common
    10·2 answers
  • What is the answer to 2x - 6 = 8 - x
    10·2 answers
  • What are the steps to solving K + 8\5 = 1
    8·2 answers
  • Kathy's internet company charges a connection fee of 0.21 cents for each call and 0.06 cents per minute for each call she made f
    11·1 answer
  • Substitute x = 2y - 2 into the expression 5x<br> (1 Point)
    11·2 answers
  • Please answer for me
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!