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IceJOKER [234]
3 years ago
5

Hii please help i’ll give brainliest!!

Mathematics
1 answer:
Valentin [98]3 years ago
3 0
Answer:


friction


lol you’re welcome
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Solve 5-2x&lt;7<br> x&lt;-1<br> x&gt;-1<br> x&lt;-12<br> X&gt;-12
Alex

5-2x

5 0
2 years ago
In 1-5, given: Two similar cylinders with heights of 8 and 5 respectively.
lana66690 [7]

Answer:

1. The ratio of their diameters is 8/5 = 8 : 5

2. The ratio of their surface area is (8/5)²

3. The ratio of their volume is (8/5)³

4.The area of the base of the larger cylinder is 128 cm²

Step-by-step explanation:

Given that the two cylinders are similar, we have;

Two cylinders are similar when the ratio of their heights is equal to the ratio of their radii

Therefore, we have;

1. The ratio of the height of the two cylinders = 8/5 = The radio of their radii = r₁/r₂

The ratio of their diameter = D₁/D₂ = 2·r₁/2·r₂ = r₁/r₂ = 8/5

The ratio of their diameters D₁/D₂ = 8/5 = 8 : 5

2. The surface area of the cylinders = 2·π·r·h + 2·π·r²

Therefore, we have;

(2·π·r₁·h₁ + 2·π·r₁²)/(2·π·r₂·h₂ + 2·π·r₂²) = (r₁·h₁ + r₁²)/(r₂·h₂ + r₂²)

h₁ = h₂ × 8/5

r₁ = r₂ × 8/5

= (8/5)²(r₂·h₂ + r₂²)/(r₂·h₂ + r₂²)  = (8/5)²

The ratio of their surface area = (8/5)²

3. The volume of the cylinder = π·r²·h

∴ The ratio of the volume = (π·r₁²·h₁)/(π·r₂²·h₂) = (8/5)³ × (π·r₂²·h₂)/(π·r₂²·h₂) = (8/5)³

The ratio of their volume = (8/5)³

4. The ratio of the area of the base of the larger cylinder to the area of the base of the smaller cylinder is (8/5)²

Therefore if the area of the base of the smaller cylinder is 50 cm², the area of the base of the larger cylinder = 50 cm² × (8/5)²  = 128 cm²

7 0
3 years ago
A play was attended by 984 persons. 75% of them were adults how many adults attend
Ierofanga [76]

Answer:

738

Step-by-step explanation:

984 ÷ 100 = 9.84

9.84 × 75 = 738

75/100

5 0
4 years ago
You decide to enter in a rowing competition. To train, you go to the boat house and begin rolling down stream. You row for the s
fredd [130]

Answer:

Time spent rowing down stream =100\ seconds

Speed of boat in still water =14\ ms^{-1}

Step-by-step explanation:

Let speed of boat in still water be = x\ ms^{-1}

Speed of current = 10\ ms^{-1}

Speed of boat down stream = \textrm{Speed of boat in still water +Speed of current}= (x+10)\ ms^{-1}

Distance rowed down stream = 2400 m

Time spent rowing down stream = \frac{Distance}{Speed}=\frac{2400}{x+10}\ s = \frac{Distance}{Speed}=\frac{2400}{x+10}\ s

Speed of boat up stream = \textrm{Speed of boat in still water -Speed of current}= (x-10)\ ms^{-1}

Distance rowed up stream = \frac{1}{6} \textrm{ of distance rowed downstream}=\frac{1}{6}\times 2400 = 400\ m

Time spent rowing up stream = \frac{Distance}{Speed}=\frac{400}{x-10}\ s

We know that,

\textrm{Time spent rowing down stream =Time spent rowing up stream}

So,

\frac{2400}{x+10}=\frac{400}{x-10}

Cross multiplying

2400(x-10)=400(x+10)

Dividing both sides by 400

\frac{2400(x-10)}{400}=\frac{400(x+10)}{400}

6(x-10)=x+10

6x-60=x+10

Adding 60 to both sides.

6x-60+60=x+10+60

6x=x+70

Subtracting both sides by x

6x-x=x+70-x

5x=70

Dividing both sides by 5.

\frac{5x}{5}=\frac{70}{5}

∴ x=14

Speed of boat in still water =14\ ms^-1

Time spent rowing down stream =\frac{2400}{14+10}=\frac{2400}{24}=100\ s

3 0
3 years ago
Using tour answer from base e, ale
Natasha2012 [34]

That’s all correct thanks

5 0
3 years ago
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