Find d^4y/dx^4 given dy/dx = x^4 + x^3 + x^2 + x
Answers
Answer:
4
Step-by-step explanation: ddx(x2+4y2)=ddx(4)
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jane walked 4/5 mile on monday on tesday she wwalked 1//2 of the distance she walked on monday how many miles she walked on tuesday a 3/4 mile b 2/5 mile c 1/5 mile d 1/2 mile
What is the intersection point of 2x+19y=3 and 1x-3=7y
I need more help please .....
Answers
s + 7 = 4t - 7 + 7
s + 7 = 4t
4 4
1/4s + 1 3/4 = t
s = 4(1/4s + 1 3/4) - 7
s = s + 7 - 7
s = s + 0
s = s
Answers
Answer:
72% your welcome lolbrhdhhdvdceuegcstditdti
52 square meters
15 square meters
104 square meters
diagram below.
Answers
area of a right triangle = ab/2 where: a and b are the legs
based on the image: a = 5 m
20 square meters = (5m * b) /2
20 sq. m * 2 = 5m * b
40 sq. m / 5m = b
8 m = b
So the image consists of 2 right triangles with a square in the middle.
Area of a square = a² = (8m)² = 64m²
Total area of the parallelogram = 2 right triangle * 20m² + 1 square *64m²
Total Area = 40m² + 64m²
Total area = 104m² ⇒ 104 square meters.
Answers
(a/b)/(c/d)=(a/b)(d/c)=(ad)/(bc)
1..
(4x/45)/(12x/25)=(4x/45)(25/12x)=(100x)/(540x)=5/27
2.
(x/9)/(3x/2)=(x/9)(2/3x)=(2x)/(27x)=2/27
3.
(7x/9)/(49/3x)=(7x/9)(3x/49)=(21x^2)/(441)=(x^2)/21
4.
(2x^2y/3yz)/(6/xz)=(2x^2y/3yz)(xz/6)=(x^3)/(9)
5. (4y-3x)/(3xy+y^2)
6. (-9x+6)/(8x+6)
7. (yx^2+y^2)/(x^2+xy^2)
8. x^-1=1/x
(x^2)/(4x+1)
Answers
Number of packs of chewing gum=x+1
Number of chocolate bars=y
The number of mints was three times the number of chocolate bars; then
x=3y
We can suggest this system of equations:
x=3y
6(x+1)+3x+10y=80
we can solve this system of equations by substitution method.
6(3y+1)+3(3y)+10y=80
18y+6+9y+10y=80
18y+9y+10y=80-6
37y=74
y=74/37
y=2
x=3y
x=3*(2)=6
Number of mints=x=6
number of packs of chewing gum=x+1=6+1=7
number of chocolate bars=y=2
Answer: the number of mints is 6, the number of packs of chewing gum is 7, and the number of chocolate bars is 2.
Answer:
He bought 13 pkgs of gum
He bought 12 mints
He bought 4 candy bars
Step-by-step explanation:
Let +a+ = number of packages of gum Terry bought
Let +b+ = number of mints
Let +c+ = number of candy bars
------------------------------
(1) +a+=+b+%2B+1+
(2) +b+=+3c+
(3) +24a+%2B+10b+%2B+35c+=+572+ ( in cents )
-------------------------------------
There are 3 equations and 3 unknowns, so it's solvable
(2) +c+=+b%2F3+
Substitute (2) into (3)
(3) +24a+%2B+10b+%2B+35%2A%28+b%2F3%29+=+572+
Substitute (1) into (3)
(3) +24%2A%28+b+%2B+1+%29+%2B+10b+%2B+35%2A%28+b%2F3+%29+=+572+
(3) +24b+%2B+24+%2B+10b+%2B+%2835%2F3%29%2Ab+=+572+
Multiply both sides by +3+
(3) +3%2A34b+%2B+3%2A24+%2B+35b+=+3%2A572+
(3) +102b+%2B+72+%2B+35b+=+1716+
(3) +137b+=+1644+
(3) +b+=+12+
and
(1) +a+=+b+%2B+1+
(1) +a+=+12+%2B+1+
(1) +a+=+13+
and
(2) +c+=+b%2F3+
(2) +c+=+12%2F3+
(2) +c+=+4+
---------------------
He bought 13 pkgs of gum
He bought 12 mints
He bought 4 candy bars
-----------------------
check:
(3) +24a+%2B+10b+%2B+35c+=+572+ ( in cents )
(3) +24%2A13+%2B+10%2A12+%2B+35%2A4+=+572+
(3) +312+%2B+120+%2B+140+=+572+
(3) +572+=+572+
OK
-x
5x - 6
5x - 30
Answers
When multiplying a term (either constant or variable) by a quantity in parentheses, we multiply each term inside the parentheses by that term. So 5 times x minus 5 times 6.