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Volgvan
3 years ago
6

Three pipes take 1 hour to water the field. How much time will it take to water the field with 4 pipes? working out please

Mathematics
1 answer:
I am Lyosha [343]3 years ago
5 0
Is it 1.3? Bc 4 divided by 3 is 1.3

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6=1-2n+5 can you please solve it
stepan [7]

Answer:

n = 0

Step-by-step explanation:

6 = 1 - 2n + 5

6 = 6 -2n

0 = -2n

n = 0

3 0
3 years ago
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A solid right pyramid has a square base with an edge length of x cm and a height of y cm.
shepuryov [24]

Answer:

B. 1/3x2y cm3

5 0
3 years ago
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Find each measurement indicated round your answers to the nearest tenth. Part 1d. NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW
valkas [14]

Answer:

see explanation

Step-by-step explanation:

Using the Sine rule in all 3 questions

(1)

\frac{a}{sinA} = \frac{b}{sinB} , substitute values , firstly calculating ∠ B

[ ∠ B = 180° - (78 + 49)° = 180° - 127° = 53° ]

\frac{a}{sin78} = \frac{18}{sin53} ( cross- multiply )

a sin53° = 18 sin78° ( divide both sides by sin53° )

a = \frac{18sin78}{sin53} ≈ 22.0 ( to the nearest tenth )

(3)

\frac{c}{sinC} = \frac{a}{sinA} , substitute values

\frac{35}{sinC} = \frac{45}{sin134} ( cross- multiply )

45 sinC = 35 sin134° ( divide both sides by 35 )

sinC = \frac{35sin134}{45} , then

∠ C = sin^{-1} ( \frac{35sin134}{45} ) ≈ 34.0° ( to the nearest tenth )

(5)

Calculate the measure of ∠ B

∠ B = 180° - (38 + 92)° = 180° - 130° = 50°

\frac{a}{sinA} = \frac{b}{sinB} , substitute values

\frac{BC}{sin38} = \frac{10}{sin50} ( cross- multiply )

BC sin50° = 10 sin38° ( divide both sides by sin50° )

BC = \frac{10sin38}{sin50} ≈ 8.0 ( to the nearest tenth )

3 0
3 years ago
Find the distance, in feet, a particle travels in its first 2 seconds of travel, if it moves according to the velocity equation
just olya [345]

Answer:

The particle will travel 6 feet in first 2 seconds.

Step-by-step explanation:

We have been given that a particle moves according to the velocity equation v(t)= 6t^2-18t+12. We are asked to find the distance that the particle will travel in its first 2 seconds.

s(t)=\int |v(t)|dt

s(t)=\int\limits^2_0 |6t^2-18t+12|dt

Now, we will eliminate the absolute value sign as:

s(t)=\int\limits^1_0 6t^2-18t+12dt+\int\limits^2_1 -6t^2+18t-12dt

s(t)=[\frac{6t^3}{3}-\frac{18t^2}{2}+12t]^1_0 +[\frac{-6t^3}{3}+\frac{18t^2}{2}-12t]^2_1

s(t)=[2t^3-9t^2+12t]^1_0 +[-2t^3+9t^2-12t]^2_1

s(2)=2(1)^3-9(1)^2+12(1)-(2(0)^3-9(0)^2+12(0))-2(2)^3+9(2)^2-12(2)-(-2(1)^3+9(1)^2-12(1))

s(2)=2-9+12-(0)-16+36-24-(-2+9-12)

s(2)=5-(0)-4-(-5)

s(2)=5-4+5

s(2)=6

Therefore, the particle will travel 6 feet in first 2 seconds.

   

 

7 0
3 years ago
I b. Mal's bank account is overdrawn by $60, which means her balance is $60. She gets $85 for her birthday and deposits it into
Mandarinka [93]
Pretty sure $145 since $60+$85=$145
6 0
3 years ago
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