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STALIN [3.7K]
3 years ago
8

PLEASE HELP THIS IS ALMOST DUE!! IF ANYONE KNOWS IT PLEASE TELL ME THANK YOU!

Mathematics
2 answers:
Allushta [10]3 years ago
8 0
X=12.35



b = (2 × area)/a
= 5.63636
c = √ a2 + √ b2
= √112 + √ 5.63636363636362
= √152.76859504132
= 12.35996
alexandr1967 [171]3 years ago
8 0
I think it D but I don’t know check
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I'm not sure where to start and how to differentiate this.
Mazyrski [523]
The picture is not clear. let me assume y = (x^4)ln(x^3)

product rule :
d f(x)g(x) = f(x) dg(x) + g(x) df(x)

dy/dx = (x^4)d[ln(x^3)/dx] + d[(x^4)/dx] ln(x^3)
= (x^4)d[ln(x^3)/dx] + 4(x^3) ln(x^3)

look at d[ln(x^3)/dx]
d[ln(x^3)/dx]
= d[ln(x^3)/dx][d(x^3)/d(x^3)]
= d[ln(x^3)/d(x^3)][d(x^3)/dx]
= [1/(x^3)][3x^2] = 3/x
... chain rule (in detail)

end up with
dy/dx = (x^4)[3/x] + 4(x^3) ln(x^3)
= x^3[3 + 4ln(x^3)]

5 0
3 years ago
Can you guys please help me!!
Karo-lina-s [1.5K]
32 it’s as simple as 4(5) + 12
7 0
3 years ago
Read 2 more answers
A crawling tractor sprinkler is located as pictured below, 100 feet South of a sidewalk. Once the water is turned on, the sprink
deff fn [24]

Answer:

  a) see below

  b) 32 minutes after turn-on

  c) 52 minutes after turn-on

  d) 20 minutes

  e) 6856.6 ft²

Step-by-step explanation:

a) We have elected to put the origin at the point where the hose crosses the south edge of the sidewalk. Units are feet. Then the sprinkler starts at (0, -100). After 1 hour, 3600 seconds, the sprinkler is 1800 inches, or 150 ft north of where it started, so stops at (0, 50).

The lines forming the sidewalk boundaries are y=0 and y=10.

__

b) Water will first strike the sidewalk when the sprinkler is 20 feet south of it, or 80 feet north of where it started. The sprinkler travels that distance in ...

  (80 ft)(12 in/ft)/(1/2 in/s)(1 min/(60 s)) = 32 min . . . time to start sprinkling sidewalk

__

c) The sprinkler has to travel to a point 130 ft north of its starting position for the water to fall north of the sidewalk. That distance is traveled in ...

  (130 ft)(2/5 min/ft) = 52 min . . . time until end of sprinkling sidewalk

Note that we have combined the scale factors in the expression of part b into one scale factor of (2/5 min/ft).

__

d) The difference of times in parts b and c is the time water falls on the sidewalk: 20 minutes.

__

e) In one hour, the sprinkler travels a distance of ...

  (60 min)(5/2 ft/min) = 150 ft

Of that distance, 10 feet is sidewalk. So, the sprinkler covers an area of grass that is a 140 ft by 40 ft rectangle and a circle of 20 ft radius. The total area of that is ...

  A = LW + πr² = (140 ft)(40 ft) +π(20 ft)² = (14+π)(400) ft² ≈ 6856.6 ft²

The area of grass watered in 1 hour is about 6856.6 ft².

5 0
4 years ago
Can someone solve 6÷2(1+2)
mamaluj [8]

Answer: 9

Step-by-step explanation:

To solve this problem, we need to use our order of operations, or PEMDAS.

Parenthesis

Exponent

Multiply

Divide

Add

Subtract

6÷2(1+2)           [parenthesis]

6÷2(3)               [multiply/divide from left to right]

3(3)                   [multiply]

9

Now, we know that the answer is 9.

8 0
4 years ago
simplify 5 square root of 11 end root minus 12 square root of 11 end root minus 2 square root of 11. negative 33 square root of
DENIUS [597]
5√11 - 12√11 - 2√11 = (5 - 12 - 2)√11 = -9√11
6 0
3 years ago
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