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3 years ago

Please help me fast

Mathematics

2 answers:

Fittoniya [83]3 years ago

6 0

Answer is 9:11 your welcome

Naddika [18.5K]3 years ago

5 0

Answer:

9:11

Step-by-step explanation:

9000-4950=4050

(refer to picture for the rest)

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P(A)=0.612,find P(-A) I need help guys

VashaNatasha [74]

Answer:

P(-A) = 0.388

Step-by-step explanation:

By the PROBABILITY POSTULATE:

P(A) + P(not A) = 1

Here, P(A) = 0.612

So, P(not A) = P(-A) = 1 - 0.612 = 0.388

⇒P(-A) = 0.388

4 0

3 years ago

Which of the following describes a situation in which the total distance a ball travel is zero meters from its starting point?

shtirl [24]

Answer:

2nd choice: The ball first bounced up to the height of 2 meters, and then falls 2 meters towards the ground

Step-by-step explanation:

3 0

3 years ago

Please solve i will give brainiest 100 point question ****** do the whole page please need to pass or i will fail its my final t

dmitriy555 [2]

Answer:

1. Find the difference between the areas.

Area of the small rectangle: $(x+2)(x+7)=x^2+7x+2x+14=x^2+9x+14$

Area of the big rectangle: $(x+9)(x+11)=x^2+11x+9x+99=x^2+20x+99$

The difference is: $11x+85$

$( x^2+20x+99)- (x^2+9x+14)=x^2+20x+99-x^2-9x-14=11x+85$

You can solve this question just by looking at the graph.

a) The height is 4 meters.

$f(d)=h=-2d^2+7d+4$

To find the height of the bleachers, we should consider the moment before the shoot, when the distance is equal to 0.

$f(0)=h=-2(0)^2+7(0)+4$

$h=4$

The height is 4 meters.

b) 9 meters.

For $d=1$

$f(1)=h=-2(1)^2+7(1)+4$

$f(1)=h=-2+7+4$

$h=9$

b) The ball travels 4 meters.

But to calculate it, it is when $h=0$

$0=-2d^2+7d+4$

Using the quadratic formula:

$d=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$d=\frac{-7 \pm \sqrt{7^2-4\left(-2\right)4}}{2\left(-2\right)}$

$d=\frac{-7\pm\sqrt{81}}{-4}$

$d=\frac{-7\pm9}{-4}$

It will give us to solutions, once it is a quadratic equation, but we are talking about a positive distance.

$d=-\frac{1}{2} \text{ or }d=4$

In this question, we have to find the area of the cylinder and the sphere.

From the information given, we have

a = 5mm and d = 5mm, therefore the radius is 2.5 mm.

The volume of a cylinder:

$V=\pi r^2h$

$V=\pi (2.5)^2 \cdot 5$

$V=31.25 \pi$

$V_{c} \approx 98.17 \text{ m}^3$

The volume of the sphere:

$V=\frac{4}{3} \pi r^2$

$V_{s} \approx 65.4 \text{ m}^3$

The volume of the capsule is approximately $163.57 \text{ m}^3$

3 0

3 years ago

What expressions are equivalent to 4x-5

Vitek1552 [10]

Answer:

2(2x-2.5)

4(x-1.25)

5(0.8x-1)

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

ZEFG and ZGFH are a linear pair, mZEFG = 3n+ 21, and mZGFH = 2n + 34. What are mZEFG and mZGFH?

marysya [2.9K]

Answer:

The answer is below

Step-by-step explanation:

∠EFG and ∠GFH are a linear pair, m∠EFG = 3n+ 21, and m∠GFH = 2n + 34. What are m∠EFG and m∠GFH?

Solution:

Two angles are said to form a linear pair if they share a base. Linear pair angles are adjacent angles formed along a line as a result of the intersection of two lines. Linear pairs are always supplementary (that is they add up to 180°).

m∠EFG = 3n + 21, m∠GFH = 2n + 34. Both angles form linear pairs, hence:

m∠EFG + m∠GFH = 180°

3n + 21 + (2n + 34) = 180

3n + 2n + 21 + 34 = 180

5n + 55 = 180

5n = 125

n = 25

Therefore, m∠EFG = 3(25) + 21 = 96°, m∠GFH = 2(25) + 34 = 84°

7 0

3 years ago

Please help me fast ​

Please help me fast