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2 years ago

8

Yr is the circundcenter of ASTU, find each measures.

2 answers:

ipn [44]2 years ago

7 0

Yes, indeed you are right! Have a nice day!

Ksju [112]2 years ago

5 0

Answer:

yes you are

Step-by-step explanation:

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Heeeeeeeeeeellllllllllllllllllllllllllllpppppppppppppppppppppp

Ira Lisetskai [31]

Answer:

A

Step-by-step explanation:

4 0

2 years ago

With a headwind, a plane traveled 390 mi in 1.5 h. Let x represent the plane’s speed (excluding wind speed) and let y represent

VMariaS [17]

The equation is ............... (it needed to be so many characters long)
390/1.5 = x - y

5 0

3 years ago

A car manufacturer is reducing the number of incidents with the transmission by issuing a voluntary recall. During week 3 of the

S_A_V [24]

Answer:

y(t) = 391 - 30t

Step-by-step explanation:

The weekly difference is 391-361 = 30

y(t) = 391 - 30t.

With t being the number of weeks.

7 0

3 years ago

Read 2 more answers

The formula T= 2pi sqrt(L/32) relates the time, T, in seconds for a pendulum with the length, L, in feet, to make one full swing

tester [92]

The length of pendulum is 2.485 feet

<h3><u><em>Solution:</em></u></h3>

Given that,

The formula T= 2pi sqrt(L/32) relates the time, T, in seconds for a pendulum with the length, L, in feet, to make one full swing back and forth

<u><em>Therefore, the given formula is:</em></u>

$T=2\pi \sqrt{\frac{L}{32} }$

We have to find the length of pendulum that makes one full swing in 1.75 seconds

So the modify the given equation to find "L"

$T=2\pi \sqrt{\frac{L}{32} }\\\\ \sqrt{\frac{L}{32} }=\frac{T}{2 \pi}\\\\\text{Taking square root on both sides }\\\\\frac{L}{32} = \frac{T^2}{4 \pi^2}\\\\L = \frac{T^2}{4 \pi^2} \times 32\\\\L = \frac{T^2}{\pi^2 } \times 8$

Substitute T = 1.75 seconds and $\pi = 3.14$

$L = \frac{1.75^2}{3.14 \times 3.14} \times 8\\\\L = \frac{3.0625}{9.8596} \times 8\\\\L = 2.485$

Thus length of pendulum is 2.485 feet approximately

7 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.

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

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

The difference is: $11x+85$

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

2.

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.

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

3.

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