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
UkoKoshka [18]
3 years ago
15

The difference of a number p and 6

Mathematics
1 answer:
frez [133]3 years ago
7 0
P-6 because you subtract the variable from the 6 
You might be interested in
Expanding log functions<br> ㏒3 (3(x+1)(x+2))
Oksanka [162]
Log(ab)=log(a)+log(b)
log_a(a)=1
so

log_3(3(x+1)(x+2))=log_3(3)+log_3(x+1)+log(x+2)=1+log_3(x+1)+log_3(x+2)
6 0
3 years ago
How do I evaluate this? there’s a picture.
PilotLPTM [1.2K]

Answer:

it is 3/3

Step-by-step explanation:

2 tan is 2/3 +cos is  1/4 +2/4

7 0
3 years ago
Evaluate 0.65 x 280 - 2/9 of 513
kondor19780726 [428]

Answer:

181.99956681

Step-by-step explanation:

Mark as brainliest

7 0
3 years ago
Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
4 years ago
Find the angle of XYZ give your answer to 1 decimal place
miskamm [114]

Answer:

66.4°

Step-by-step explanation:

To find the angle XYZ, we are to use sine rule. For this, we have to first find ∠Z.

Given that: ∠X = 90° (right angle), XY = 6 cm, YZ = 15 cm. Hence:

\frac{sin(Z)}{XY}=\frac{sin(X)}{YZ}\\\\Substituting:\\\\\frac{sin(Z)}{6} =\frac{sin(90)}{15}   \\\\sin(Z)=\frac{sin(90)*6}{15}\\\\sin(Z)=0.4\\\\Z=sin^{-1}(0.4)\\\\Z=23.6^o

∠X + ∠Y + ∠Z = 180° (sum of angles in a triangle)

90 + ∠Y + 23.6 = 180

113.6 + ∠Y = 180

∠Y = 180 - 113.6

∠Y = 66.4°

∠Y = ∠XYZ = 66.4°

3 0
3 years ago
Other questions:
  • Number 18 idk what to do help plz and fast
    11·1 answer
  • Write an expression to show how you could calculate the total amount David paid.
    15·2 answers
  • Q: If 42 + 75 = 165<br> what does 64 + 56<br> equal to?
    7·2 answers
  • The phone company V-Mobile charges $100 a month for their family plan plus an additional $30 per phone line on the plan. Write a
    14·1 answer
  • Correct answer gets brainliest! :)
    6·2 answers
  • If you know how to solve this, Please answer it. Thank You
    14·2 answers
  • Simplify: 299.081+59.26-95.32-365.195
    5·1 answer
  • Joy and Lane share a 24-ounce bucket of clay. By the end of the week, Joy has used
    11·1 answer
  • The four Fifth Grade classes at Smith Elementary School are having a competition to collect Box Tops. Below is the information a
    15·1 answer
  • Help i give 5 stars and a crown​
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!