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
user100 [1]
3 years ago
12

What is the volume? Please answer asap.

Mathematics
1 answer:
ICE Princess25 [194]3 years ago
3 0

Answer:

20

Step-by-step explanation:

You might be interested in
9+10=????????????????????????
givi [52]

Answer:

=19

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
First person with a answer is marked as the Brainliest answer
Svet_ta [14]

Answer:

8

Step-by-step explanation:

look at the graph

5 0
3 years ago
Read 2 more answers
If Kevin wants a 90 average in math after 5 tests and his first 4 tests are 76, 92, 89 and 97 what does he need on the fifth tes
Keith_Richards [23]
You need to understand that you're solving for the average, which you already know: 90. Since you know the values of the first three exams, and you know what your final value needs to be, just set up the problem like you would any time you're averaging something.
Solving for the average is simple:
Add up all of the exam scores and divide that number by the number of exams you took.
(87 + 88 + 92) / 3 = your average if you didn't count that fourth exam.
Since you know you have that fourth exam, just substitute it into the total value as an unknown, X:
(87 + 88 + 92 + X) / 4 = 90
Now you need to solve for X, the unknown:
87
+
88
+
92
+
X
4
(4) = 90 (4)
Multiplying for four on each side cancels out the fraction.
So now you have:
87 + 88 + 92 + X = 360
This can be simplified as:
267 + X = 360
Negating the 267 on each side will isolate the X value, and give you your final answer:
X = 93
Now that you have an answer, ask yourself, "does it make sense?"
I say that it does, because there were two tests that were below average, and one that was just slightly above average. So, it makes sense that you'd want to have a higher-ish test score on the fourth exam.
4 0
4 years ago
Read 2 more answers
What is the area of a 30 degree sector of a circle with area 96 pi meters ^2?
iren2701 [21]
\bf \textit{area of a circle}\\\\
A=\pi r^2~~
\begin{cases}
r=radius\\
-----\\
A=96\pi 
\end{cases}\implies 96\pi =\pi r^2
\\\\\\
\cfrac{96\underline{\pi }}{\underline{\pi }}=r^2\implies 96=r^2\implies \boxed{\sqrt{96}=r}\\\\
-------------------------------

\bf \textit{area of a sector of a circle}\\\\
A=\cfrac{\theta \pi r^2}{360}~~
\begin{cases}
r=radius\\
\theta = angle~in\\
\qquad degrees\\
------\\
\theta =30\\
r=\boxed{\sqrt{96}}
\end{cases}\implies A=\cfrac{(30)(\pi )(\sqrt{96})^2}{360}
\\\\\\
A=\cfrac{(30)(\pi )(96)}{360}\implies A=8\pi
4 0
4 years ago
What does tanx sinx (cos^2x cscx- sinx) simplify to
Elan Coil [88]
Tanx sinx (cos^2x cscx- sinx)


=(sinx/cosx)((cos^2x cscx- sinx)

=(sinx/cosx)((cos^2x 1/sinx- sinx)

=(sinx/cosx)((cos^²x - sin²x)] / sinx

=(sinx/cosx)((cos^²x - sin²x)] / sinx

=(1/cosx)((cos^²x - sin²x)]

=(1/cosx)(2cos²x====> = 2 cos x]


7 0
4 years ago
Read 2 more answers
Other questions:
  • Solve for x: 0.75x-0.25=3.5
    6·1 answer
  • -2 (3x - 4)= -6x - 8 true or false
    15·2 answers
  • Please help you don't have to show work
    10·1 answer
  • 1. What is the factored form of 5x2 - 8x – 4?
    5·2 answers
  • Solve using the quadratic formula. (Identify a, b, and c. Write the formula. Show the substitution step. Show all work.)
    9·2 answers
  • A bag contains 7 red marbles, 3 blue marbles and 6 green marbles. If two
    9·1 answer
  • Find the mean of 9,5,10 graphically will mark Brainliest if correct ( no files or links)​
    7·2 answers
  • The floor of the cafeteria storage room is being tiled. The room is shaped like a parallelogram. The
    12·2 answers
  • Which of the following have eight faces
    7·1 answer
  • Can someone please help me with part c of this question? Thank you!
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!