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3 years ago

a radius is an angle that connects any points on the circle to the center of that circle ture or false

Mathematics

1 answer:

evablogger [386]3 years ago

7 0

The answer is false because radius is a segment that connects any one point on a circle to the center point.

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HELPP FASTT!!

joja [24]

Answer:

Those questions are from a story or something like that, so if you dont send the story we can not help you :(

8 0

3 years ago

Read 2 more answers

3. The measure of side c is 110 inches and

Pavel [41]

C is it I think let me know if I’m wrong

4 0

3 years ago

Instructions: Find the missing side. Round your answer to the nearest tenth.<br> 28°<br> х<br> 23

Illusion [34]

Answer:

x=49.0

Step-by-step explanation:

Hi there!

In this right triangle, we're given an angle and the opposing side, and we're trying to solve for the length of the hypotenuse.

In this situation, we can use the sine ratio:

$sin\theta=\frac{opp}{hyp}$

Plug in the known information

$sin(28)=\frac{23}{x}\\x=\frac{23}{sin(28)}\\x=49.0$

Therefore, the length of the missing side is 49.0 units when rounded to the nearest tenth.

I hope this helps!

8 0

3 years ago

First-order linear differential equations

kkurt [141]

Answer:

$(1)\ logy\ =\ -sint\ +\ c$

$(2)\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Step-by-step explanation:

1. Given differential equation is

$\dfrac{dy}{dt}+ycost = 0$

$=>\ \dfrac{dy}{dt}\ =\ -ycost$

$=>\ \dfrac{dy}{y}\ =\ -cost dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{y}}\ =\ \int{-cost\ dt}$

$=>\ logy\ =\ -sint\ +\ c$

Hence, the solution of given differential equation can be given by

$logy\ =\ -sint\ +\ c.$

2. Given differential equation,

$\dfrac{dy}{dt}\ -\ 2ty\ =\ t$

$=>\ \dfrac{dy}{dt}\ =\ t\ +\ 2ty$

$=>\ \dfrac{dy}{dt}\ =\ 2t(y+\dfrac{1}{2})$

$=>\ \dfrac{dy}{(y+\dfrac{1}{2})}\ =\ 2t dt$

On integrating both sides, we will have

$\int{\dfrac{dy}{(y+\dfrac{1}{2})}}\ =\ \int{2t dt}$

$=>\ log(y+\dfrac{1}{2})\ =\ 2.\dfrac{t^2}{2}\ + c$

$=>\ log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

Hence, the solution of given differential equation is

$log(y+\dfrac{1}{2})\ =\ t^2\ +\ c$

8 0

4 years ago

A heavy rope, 30 ft long, weighs 0.4 lb/ft and hangs over the edge of a building 80 ft high. Approximate the required work by a

gizmo_the_mogwai [7]

Answer:

180 fb*lb

45 ft*lb

Step-by-step explanation:

We have that the work is equal to:

W = F * d

but when the force is constant and in this case, it is changing.

therefore it would be:

$W = \int\limits^b_ a {F(x)} \, dx$

Where a = 0 and b = 30.

F (x) = 0.4 * x

Therefore, we replace and we would be left with:

$W = \int\limits^b_a {0.4*x} \, dx$

We integrate and we have:

W = 0.4 / 2 * x ^ 2

W = 0.2 * (x ^ 2) from 0 to 30, we replace:

W = 0.2 * (30 ^ 2) - 0.2 * (0 ^ 2)

W = 180 ft * lb

Now in the second part it is the same, but the integral would be from 0 to 15.

we replace:

W = 0.2 * (15 ^ 2) - 0.2 * (0 ^ 2)

W = 45 ft * lb

6 0

3 years ago