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2 years ago

Pleaseee help answer correctly !!!!!!!!!!!!!!!! Will mark Brianliest !!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Naya [18.7K]2 years ago

5 0

Answer:

540

Step-by-step explanation:

(n-2)×180

(5-2)×180

3×180

540

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Dan draws a circle with a radius of 8 inches. Question 1 Part A: What is the circumference of Dan's circle, inches? If needed, r

geniusboy [140]

Answer:

The circumference of Dan's circle is approximately 50.27 inches.

Step-by-step explanation:

The length of the circumference ( $s$ ), measured in inches, is determined by the formula:

$s = 2\pi\cdot r$ (1)

Where $r$ is the radius, measured in inches.

If we know that $r = 8\,in$ , then the length of the circumference is:

$s = 2\pi\cdot (8\,in)$

$s \approx 50.27\,in$

The circumference of Dan's circle is approximately 50.27 inches.

8 0

3 years ago

A ship's sonar finds that the angle of depression to a wreck on the bottom of the ocean is 15.20. If a point on

Alexxx [7]

Answer:

We have

tan 12.5 = 60 / adj rearrange as

adj = 60 / tan 12.5 = about 270.64 m

Step-by-step explanation:

Mark me brainllest

3 0

2 years ago

Simplify. 32 + 16 + 8 A) 4 + 6 2 B) 6 + 4 2 C) 10 + 2 D) 10 2

Ann [662]

Answer:

its 56 if you simplify 32 + 16 + 8

6 0

3 years ago

Really need help please help me please please

Maru [420]

Answer:

y = 3/4x + 3

Step-by-step explanation:

you should only have y in one side and everything in other side of the equation. so you will first have -4y = -3x -12 , then you'll have to divide the right part of the equation by -4 . then you will have y = 3/4x + 3

3 0

3 years ago

assume x and y are both differentiable functions of t. find dx/dt given x=-1 and dy/dt=8 for the relation: 4x^2+3y^3=28

melomori [17]

Given:

x and y are both differentiable functions of t.

$4x^2+3y^3=28$

$x=-1\text{ and }\dfrac{dy}{dt}=8$

To find:

The value of $\dfrac{dx}{dt}$ .

Solution:

We have,

$4x^2+3y^3=28$ ...(i)

At x=-1,

$4(-1)^2+3y^3=28$

$4+3y^3=28$

$3y^3=28-4$

$3y^3=24$

Divide both sides by 3.

$y^3=8$

Taking cube root on both sides.

$y=2$

So, y=2 at x=-1.

Differentiate (i) with respect to t.

$4(2x\dfrac{dx}{dt})+3(3y^2\dfrac{dy}{dt})=0$

Putting x=-1, y=2 and $\dfrac{dy}{dt}=8$ , we get

$4(2(-1)\dfrac{dx}{dt})+3(3(2)^28)=0$

$-8\dfrac{dx}{dt}+9(4)(8)=0$

$-8(\dfrac{dx}{dt}-9(4))=0$

Divide both sides by -8.

$\dfrac{dx}{dt}-36=0$

$\dfrac{dx}{dt}=36$

Therefore, the value of $4x^2+3y^3=28$ is 36.

6 0

2 years ago